Exercises: Convergence criteria for series – "Math for Non-Geeks"

Application of convergence criteria

Exercise (Convergence proof training 1)

Investigate, whether the following series are divergent, convergent or even absolutely convergent.

1. $\sum _{k=1}^{\mathrm{\infty }}\frac{(k^k{)}^2}{k^{(k^2)}}$
2. $\sum _{k=1}^{\mathrm{\infty }}\frac{(k!{)}^2}{(2k)!}$
3. $\sum _{k=1}^{\mathrm{\infty }}\frac{k}{1+k^2}$
4. $\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{k}{k+1}\right)}^k$
5. $\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{k}{k+1}\right)}^{k^2}$
6. $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k(\sqrt{k+1}-\sqrt{k})$
7. $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k}{\sqrt[k]{k}}$
8. $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k(1+\frac{1}{k}{)}^k}{k}$

Solution (Convergence proof training 1)

1. Root test: The $k$ in the exponents is tedious. We can remove it from one exponent by taking the $k$-th root: $\begin{array}{cc}\sqrt[k]{|a_k|}& =\sqrt[k]{\left|\frac{(k^k{)}^2}{k^{(k^2)}}\right|}=\frac{\sqrt[k]{k^{2k}}}{\sqrt[k]{k^{(k^2)}}}=\frac{k^2}{k^k}=\frac{1}{k^{k-2}}⟶0<1\end{array}$ The root test now tells us that the series converges absolutely. Intuitively, the exponent $(\cdot {)}^{k^2}$ produces a much faster growth than $(\cdot {)}^{2k}$ . So fast that $|a_k|$ decays faster than a geometric series $b_k=q^k,0\le q<1$.

2. Ratio test: By which factor do the sequence elements $|a_k|$ increase, if we go from $|a_k|$ to $|a_{k+1}|$? If that factor is lower than some constant $q<1$, we have absolute convergence:

$\begin{array}{cc}\left|\frac{a_{k+1}}{a_k}\right|& =\frac{\frac{((k+1)!{)}^2}{(2k+2)!}}{\frac{(k!{)}^2}{(2k)!}}=\frac{((k+1)!{)}^2(2k)!}{(k!{)}^2(2k+2)!}=\frac{k!k!(k+1{)}^2(2k)!}{k!k!(2k)!(2k+1)(2k+2)}=\frac{k^2+2k+1}{4k^2+6k+2}\\ & =\frac{1+\frac{2}{k}+\frac{1}{k^2}}{4+\frac{6}{k}+\frac{2}{k^2}}⟶\frac{1}{4}<1\end{array}$

So the increase factor is smaller than $q=1/4$ and the series converges absolutely.

3. Direct comparison test: For large $k$, the series elements essentially behave like $a_k\approx \frac{k}{k^2}=\frac{1}{k}$, which is a divergent harmonic series. So we suspect that this series also diverges. In fact, the $+1$ in the denominator makes $a_k$ smaller than $\frac{1}{k}$. But for $k\to \mathrm{\infty }$, $a_k$ will approach it arbitrarily close and get larger that $c\cdot \frac{1}{k}$ for any $0\le c<1$. For convenience, we choose $c=1/2$ and get that eventually:

• $|a_k|=\frac{k}{1+k^2}\ge \frac{k}{k^2+k^2}=\frac{k}{2k^2}=\frac{1}{2k}$
• But the harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{2k}$ diverges.

The series $a_k$ is "even larger" and therefore also diverges.

4. Term test: What happens for $k\to \mathrm{\infty }$? Actually, the series elements get close to a constant:

$a_k={\left(\frac{k}{k+1}\right)}^k=\frac{1}{{\left(\frac{k+1}{k}\right)}^k}=\frac{1}{{(1+\frac{1}{k})}^k}\to \frac{1}{e}\ne 0.$

The series would intuitively evaluate to $\sum _k{a}_k\approx \frac{1}{e}\mathrm{\infty }=\mathrm{\infty }$ and is in mathematical words divergent by the term test.

5. Root test: For large $k$, we have seen in 4., that ${\left(\frac{k}{k+1}\right)}^k$ goes like $\frac{1}{e}$. So ${\left(\frac{k}{k+1}\right)}^{k^2}$ should behave like $\frac{1}{e^k}$, which is a geometric series and hence convergent. We can verify this convergence by the root criterion:

$\sqrt[k]{|a_k|}={\left(\frac{k}{k+1}\right)}^k=\frac{1}{{\left(\frac{k+1}{k}\right)}^k}=\frac{1}{{(1+\frac{1}{k})}^k}\to \frac{1}{e}<1.$

Hence, the series converges absolutely.

6. Alternating series test:

The $(-1{)}^k$ suggests that we have to deal with an alternating series. And since $\sqrt{k+1}-\sqrt{k}$ is always positive, this is indeed true. Further, the two square roots are moving very close together for high $k$. For instance $\sqrt{1}-\sqrt{0}=1$, but $\sqrt{101}-\sqrt{100}\approx 0.05$ and $\sqrt{1000001}-\sqrt{1000000}\approx 0.0005$ (you may verify this yourself by taking the first-order Taylor approximation or a pocket calculator). We therefore suspect $\sqrt{k+1}-\sqrt{k}$ to be monotonically decreasing, which we will prove in the following:

$\sqrt{k+1}-\sqrt{k}=\frac{(\sqrt{k+1}-\sqrt{k})(\sqrt{k+1}+\sqrt{k})}{\sqrt{k+1}+\sqrt{k}}=\frac{k+1-k}{\sqrt{k+1}+\sqrt{k}}=\frac{1}{\sqrt{k+1}+\sqrt{k}}$

So

• $a_k=\frac{1}{\sqrt{k+1}+\sqrt{k}}$ is indeed monotonically decreasing, since $\sqrt{k+2}+\sqrt{k+1}\ge \sqrt{k+1}+\sqrt{k}⟺\frac{1}{\sqrt{k+2}+\sqrt{k+1}}\le \frac{1}{\sqrt{k+1}+\sqrt{k}}\mathrm{\forall }k\in \mathbb{N}$
• $(a_k)$ is a null sequence, since $0\le a_k\le \frac{1}{2\sqrt{k}}=\frac{1}{2}\frac{1}{\sqrt{k}}\to 0$.

So the series converges.

However, the series does not converge absolutely, since we can write $\sum _k|a_k|$ as a divergent telescoping series

$\sum _{k=1}^{\mathrm{\infty }}(\sqrt{k+1}-\sqrt{k})=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n(\sqrt{k+1}-\sqrt{k})\underset{\text{sum}}{\stackrel{\text{telescoping}}{=}}\underset{n\to \mathrm{\infty }}{lim}\sqrt{n+1}-1=\mathrm{\infty }$

7. Term test: For large $k$, the expression $\sqrt[k]{k}$ converges to 1. So we expect the absolute value of $a_k=\frac{(-1{)}^k}{\sqrt[k]{k}}$ to converge to 1 and the series should diverge by means of the term test. Now, every second element $a_k$ is negative, so we have to restrict to a subsequence of positive elements: $a_k=\frac{(-1{)}^k}{\sqrt[k]{k}}\Rightarrow a_{2l}=\frac{1}{\sqrt[2l]{2l}}⟶\frac{1}{1}=1\text{ (since }\sqrt[k]{k}\to 1\text{)}$ Indeed, the subsequence $(a_{2l})$ of $(a_k)$ doesn't converge to 0, so $(a_k)$ cannot be a null sequence and by means of the term test, the series $\sum _k{a}_k$ is divergent.

Note: The $(-1{)}^k$ may tempt one to assume that the alternating series test is a good tool to use. However, the alternating series test can only prove convergence and as the series is divergent, any attempt using it bound to fail! Indeed, $(b_k)=(\frac{1}{\sqrt[k]{k}})$ is not a null sequence, which is why the alternating series test doesn't work.

8. alternating series test: The ${\left(\frac{k+1}{k}\right)}^k$ converges to $e$, so the series elements behave like $a_k\approx (-1{)}^k\frac{e}{k}$ for large $k$. This is an alternating harmonic series, so we suspect $\sum _k{a}_k$ to converge, but not absolutely. We prove this using the alternating series test: For $a_k=\frac{(1+\frac{1}{k}{)}^k}{k}={\left(\frac{k+1}{k}\right)}^k\frac{1}{k}$ there is

• $\frac{a_{k+1}}{a_k}=\frac{(\frac{k+2}{k+1}{)}^{k+1}\frac{1}{k+1}}{(\frac{k+1}{k}{)}^k\frac{1}{k}}=\frac{(\frac{k+2}{k+1}{)}^{k+1}}{(\frac{k+1}{k}{)}^k\frac{k+1}{k}}={\left(\frac{\frac{k+2}{k+1}}{\frac{k+1}{k}}\right)}^{k+1}={\left(\frac{(k+2)k}{(k+1{)}^2}\right)}^{k+1}={\left(\frac{k^2+2k}{k^2+2k+1}\right)}^{k+1}\le 1$ , so $a_k$ is monotonically decreasing.
• $a_k=\frac{(1+\frac{1}{k}{)}^k}{k}\to 0$, since $(1+\frac{1}{k}{)}^k\to e$. So $(a_k)$ is a null sequence.

Therefore, the series converges.

To show that it does not converge absolutely, we compare $|a_k|$ to a smaller but still divergent harmonic series (direct comparison test):

• $|a_k|=\frac{(1+\frac{1}{k}{)}^k}{k}\ge \frac{(1+\frac{1}{1}{)}^1}{k}=\frac{2}{k}$, since ${\tilde{a}}_k=(1+\frac{1}{k}{)}^k$ is monotonically increasing.
• $\sum _{k=1}^{\mathrm{\infty }}\frac{2}{k}$ still diverges. (harmonic series)

And hence, the series of absolute values $\sum _{k=1}^{\mathrm{\infty }}|a_k|=\sum _{k=1}^{\mathrm{\infty }}\frac{(1+\frac{1}{k}{)}^k}{k}$ diverges.

Exercise (Convergence proof training 2)

Investigate, whether the following series are divergent, convergent or even absolutely convergent.

1. $\sum _{k=2}^{\mathrm{\infty }}\frac{\ln k}{k^3}$
2. $\sum _{k=2}^{\mathrm{\infty }}\frac{1}{\ln k}$
3. $\sum _{k=1}^{\mathrm{\infty }}{k}^4\exp (-k)$
4. $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\cos (\frac{1}{k})$
5. $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\sin (\frac{1}{k})$
6. $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\cosh k}$
7. $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{\sin k}{k^2}$

Solution (Convergence proof training 2)

1. Direct comparison test: It is useful to know that $\ln k$ eventually grows slower than any of the polynomials $k,k^{0.5},k^{0.1},k^{0.01},\mathrm{...}$, i.e. any $k^{\epsilon }$ with $\epsilon >0$. We can generously take $ϵ=1$ and estimate $\ln k\le k$ :

• $|a_k|=\frac{\ln k}{k^3}\le \frac{k}{k^3}=\frac{1}{k^2}$
• $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}=\zeta (2)=\frac{{\pi }^2}{6}<\mathrm{\infty }$

So the series elements grow slower than $\frac{1}{k^2}$ and the series converges absolutely.

Note: For $\epsilon >0$ , there is $\ln k\le c_{\epsilon }{k}^{\epsilon }$ and we could even get $|a_k|\le \frac{1}{k^{3-\epsilon }}$.

2. Direct comparison test: Since $\ln k\le k$, the series elements grow slower than

• $a_k=|a_k|=\frac{1}{\ln k}\ge \frac{1}{k}$
• but $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ diverges (harmonic series)

So the series diverges.

3. Ratio test: We write $a_k=k^4\exp (-k)=\frac{k^4}{e^k}$ . Both $k^4$ and $e^k$ grow fast at high $k$. But eventually, the exponential $e^k$ "wins" against all polynomial functions ,such that $a_k\to 0$ and even $\sum _k{a}_k<\mathrm{\infty }$. Mathematically, this can be proven by considering the ratio between two subsequent elements: $\frac{|a_{k+1}|}{|a_k|}=\frac{\frac{(k+1{)}^4}{e^{k+1}}}{\frac{k^4}{e^k}}=\frac{(k+1{)}^4}{k^4}\frac{e^k}{e^{k+1}}={(1+\frac{1}{k})}^4\frac{1}{e}\to \frac{1}{e}<1$ So the series converges absolutely.

Note: We could have done the same proof with ${\tilde{a}}_k=\frac{k^n}{e^k}$ for any $n\ge 1$ (not just $n=4$). The exponential always "wins" against the polynomial.

Alternative: Root est: $\sqrt[k]{|a_k|}=\sqrt[k]{\frac{k^4}{e^k}}=\frac{\sqrt[k]{k^4}}{\sqrt[k]{e^k}}=\frac{{\sqrt[k]{k}}^4}{e}\to \frac{1}{e}<1$ So the series converges absolutely.

4. Term test: For $a_k=(-1{)}^k\cos (\frac{1}{k})$ , the $\cos (\frac{1}{k})$ approaches 1 as $k\to \mathrm{\infty }$ . We select a subsequence consisting of only the positive entries: $a_{2l}=\cos (\frac{1}{2l})\to \cos (0)=1\ne 0$ Therefore, $(a_k)$ cannot be a null sequence and the corresponding series diverges.

Note: The $(-1{)}^{k+1}$ may trick you into taking the alternating series test for proving that the series converges. However, this doesn't work since $b_k=\cos (\frac{1}{k})$ is not a null sequence!

5. Alternating series test: This time, $\sin (\frac{1}{2l})\to 0$ , as $k\to \mathrm{\infty }$ . So we have a $(-1{)}^{k+1}$ in front of a null sequence and may try the alternating series test:

• $b_{k+1}=\sin (\frac{1}{k+1})\le \sin (\frac{1}{k})=b_k$, since $\sin$ is monotonically decreasing. Therefore, $(b_k)$ is also monotonically decreasing.
• $b_k=\sin (\frac{1}{k})\to \sin (0)=0$, since $\sin$ is continuous. Therefore, $(b_k)$ is a null sequence.

By the alternating series test, the series converges.

Note: The series does not converge absolutely, since for large $k$, there is $b_k=\sin (\frac{1}{k})>\frac{1}{2k}$. And ${\tilde{b}}_k=\frac{1}{2k}$ is a divergent harmonic series.

6. Direct comparison test: The $\cosh$ can be written as $a_k=\frac{1}{\cosh k}=\frac{1}{\frac{e^k+e^{-k}}{2}}=\frac{2}{e^k+e^{-k}}$. So it decays like $\frac{2}{e^k}$ (a geometric series). We compare $a_k$ to this geometric series

• $|a_k|=\frac{2}{e^k+e^{-k}}\le \frac{2}{e^k}=2\cdot \frac{1}{e^k}$, since $e^{-k}>0$.
• $\sum _{k=1}^{\mathrm{\infty }}2\cdot \frac{1}{e^k}=2\cdot \sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{e}\right)}^k=\frac{2e}{1-\frac{1}{e}}<\mathrm{\infty }$ (geometric series)

Therefore, the series converges absolutely.

7. Direct comparison test: The $\sin$ is oscillating with amplitude $\le 1$ (therefore staying bounded) and the $\frac{1}{k^2}$ decays fast enough to get convergence. We can hence compare:

• $|a_k|=\frac{|\sin k|}{k^2}\le \frac{1}{k^2}$
• $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}=\zeta (2)=\frac{{\pi }^2}{6}<\mathrm{\infty }$

So the series converges absolutely.

Note: Even though the series is oscillating, the alternating series test doesn't work here, since $b_k=\frac{\sin k}{k^2}$ is not monotonically decreasing!

Exercise (Series depending on parameters)

The following series depend on a parameter $x\in R$ . For some $x$ , they will converge, for others not. Your job is to investigate for which $x\in ℝ$ , the series converges (absolutely) or diverges:

1. $\sum _{k=1}^{\mathrm{\infty }}\frac{x^k}{(2k)!}$
2. $\sum _{k=1}^{\mathrm{\infty }}\frac{x^k}{k^2}$
3. $\sum _{k=1}^{\mathrm{\infty }}\sqrt{k}{x}^k$
4. $\sum _{k=1}^{\mathrm{\infty }}\frac{x^k}{k}$

Solution (Series depending on parameters)

1. Ratio test: Intuitively, a factorial $k!$ grows faster than any exponential $x^k$ in a way that not just $\frac{x^k}{k!}\to 0$ , but even the series $\sum _k\frac{x^k}{k!}$ converges. This should also hold true if we replace $k!$ by the larger expression $(2k)!$ . We can verify this mathematically by the ratio test:

$\frac{|a_{k+1}|}{|a_k|}=\frac{\frac{|x{|}^{k+1}}{(2k+2)!}}{\frac{|x{|}^k}{(2k)!}}=\frac{|x{|}^{k+1}}{|x{|}^k}\cdot \frac{(2k)!}{(2k+2)!}=\frac{|x|}{(2k+1)(2k+2)}=\frac{|x|}{4k^2+6k+2}\to 0<1$

So the series is absolutely convergent for all $x\in ℝ$.

2. Case distinction:

Case 1: $|x|\le 1$

The series elements are smaller than $|a_k|=\frac{|x{|}^k}{k^2}\le \frac{1}{k^2}$ , which is known to be a convergent series: $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{k^2}=\zeta (2)=\frac{{\pi }^2}{6}<\mathrm{\infty }$

So by the direct comparison test, the series converges absolutely.

Case 2: $|x|>1$

In this case, we expect the $x^k$ to "explode exponentially" in $x$, while $k^2$ is growing only polynomially (and hence slower). So the series $\frac{x^k}{k^2}$ should converge. Mathematically, this can be verified by the root test: $\sqrt[k]{|a_k|}=\sqrt[k]{\frac{|x{|}^k}{k^2}}=\frac{|x|}{\sqrt[k]{k^2}}=\frac{|x|}{{\sqrt[k]{k}}^2}\to |x|>1$, since $\sqrt[k]{k}\to 1$

So the series diverges.

3. Case distinction:

We now distinguish two cases:

Case 1: $|x|<1$

In that case, the $x^k$ decays exponentially, and therefore "wins" against the polynomially increasing $\sqrt{k}=k^{1/2}$. We use the ratio test to verify this:

$\frac{|a_{k+1}|}{|a_k|}=\frac{\sqrt{k+1}|x{|}^{k+1}}{\sqrt{k}|x{|}^k}=\sqrt{1+\frac{1}{k}}|x|\to |x|<1$

So indeed, the series diverges absolutely.

Case 2: $|x|\ge 1$

Here, the $x^k$ is constant ($|x|=1$) or exponentially increasing. Multiplying it by $\sqrt{k}$ we get sequence elements which are increasing and hence bigger than a constant. This suggests using the term test: $|a_k|=\sqrt{k}|x{|}^k\ge \sqrt{k}\to \mathrm{\infty }$. So $(a_k)$ is not a null sequence.

Therefore, the series diverges.

4. Case distinction:

Case 1: $|x|<1$

Here, $|a_k|=\frac{|x{|}^k}{k}\le |x{|}^k$ and $\sum _{n=1}^{\mathrm{\infty }}|x{|}^k<\mathrm{\infty }$ (geometric series)

So by the direct comparison test, the series converges absolutely.

Case 2: $x=1$

$\sum _{k=1}^{\mathrm{\infty }}\frac{1^k}{k}=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ , which is a divergent harmonic series.

Case 3: $x=-1$

$\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k}{k}$ is an alternating harmonic series, which converges but not absolutely, by the alternating series test.

Case 4: $|x|>1$

We have $|a_k|=\frac{|x{|}^k}{k}\ge \frac{1}{k}$, where the geometric series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ diverges (direct comparison test).

Note: The cases $|x|<1$ and $|x|>1$ can also be treated by the root test or the ratio test.

Exercise (Direct comparison in asymptotic behaviour)

Investigate, whether the series

$\sum _{k=1}^{\mathrm{\infty }}\frac{7k^7(1+\frac{1}{k})(k^3-1)}{(k^3+2)(2k^5+1)(k^2+8{)}^2}$

converges or diverges.

Solution (Direct comparison in asymptotic behaviour)

We take a look at the asymptotic behaviour for $k\to \mathrm{\infty }$:

$a_k=\frac{7k^{10}(1+\frac{1}{k})(1-\frac{1}{k^3})}{2k^{12}(1+\frac{2}{k^3})(1+\frac{1}{2k^5})(1+\frac{8}{k^2}{)}^2}=\underset{=\frac{7}{2}\frac{1}{k^2}}{\underbrace{\frac{7k^{10}}{2k^{12}}}}\underset{\to 1\text{ for }k\to \mathrm{\infty }}{\underbrace{\frac{(1+\frac{1}{k})(1-\frac{1}{k^3})}{(1+\frac{2}{k^3})(1+\frac{1}{2k^5})(1+\frac{8}{k^2}{)}^2}}}$

So it might be useful to compare $a_k$ with a series scaling like $b_k=\frac{1}{k^2}$. In fact,

$\underset{k\to \mathrm{\infty }}{lim}\frac{a_k}{b_k}=\underset{k\to \mathrm{\infty }}{lim}\frac{7}{2}\frac{(1+\frac{1}{k})(1-\frac{1}{k^3})}{(1+\frac{2}{k^3})(1+\frac{1}{2k^5})(1+\frac{8}{k^2}{)}^2}=\frac{7}{2}\frac{(1+0)(1-0)}{(1+0)(1+0)(1+0{)}^2}=\frac{7}{2}$

This convergence tells us that for high $k$ , there must be a $M\in \mathbb{N}$ with

$|a_k|\le 2\cdot \frac{7}{2}|b_k|=7b_k$ for all $k\ge M$

So we can compare $|a_k|$ to $7b_k=\frac{7}{k^2}$ . Since the series $\sum _{k=1}^{\mathrm{\infty }}\frac{7}{k^2}=7\frac{{\pi }^2}{6}$converges, we know by the direct comparison test that also $\sum _{k=1}^{\mathrm{\infty }}$ converges absolutely.

Term test - strengthened version

Exercise (Term test - strengthened version)

The term test tells us that if $(a_n{)}_{n\in \mathbb{N}}$ is monotonically decreasing and $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges, then $(a_n{)}_{n\in \mathbb{N}}$ is a null sequence. Prove that in this case, even $(n{a}_n{)}_{n\in \mathbb{N}}$ must be a null sequence. (This means that it suffices to show that $(n{a}_n{)}_{n\in \mathbb{N}}$ is not a null sequence in order to conclude that $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.)

Solution (Term test - strengthened version)

The intuition behind this test is that $(n{a}_n{)}_{n\in \mathbb{N}}$ being not a null sequence means that $(a_n)$ decays slower than or as fast as a harmonic sequence $b_n=\frac{1}{n}$, and $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ diverges. A formal proof can be done by using the Cauchy criterion: Assume that $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges. Then, the sequence of partial sums $(s_k{)}_{k\in \mathbb{N}}=\sum _{n=1}^k{a}_n$ is a Cauchy sequence. We will first use this fact to bound all even elements of $(a_n{)}_{n\in \mathbb{N}}$:

Proof step: $(2k{a}_{2k}{)}_{k\in \mathbb{N}}$ is a null sequence

Since $(s_k{)}_{k\in \mathbb{N}}=\sum _{n=1}^k{a}_n$ is a Cauchy sequence, there must be an $ϵ>0$ and an $N\in \mathbb{N}$, such that for all $k+1\ge N$ there is

$|s_n-s_k|=\sum _{n=k+1}^{2k}{a}_n=a_{k+1}+a_{k+2}+\dots +a_{2k-1}+a_{2k}<ϵ$

Hence, for all $k+1\ge N$:

$\begin{array}{cc}\frac{1}{2}\cdot 2k{a}_{2k}& =k{a}_{2k}\\ & =a_{2k}+a_{2k}+\dots +a_{2k}+a_{2k}\\ & \downarrow (a_n)\text{ decreases monotonically}\\ & \le a_{k+1}+a_{k+2}+\dots +a_{2k-1}+a_{2k}<ϵ\end{array}$

So we know that $(k{a}_{2k})$ and hence also $(2k{a}_{2k})$ eventually falls below each $ϵ>0$ , so they are null sequences.

Now we are still missing the odd numbers:

Proof step: $((2k-1)a_{2k-1}{)}_{k\in \mathbb{N}}$ is a null sequence

The argumentation is the same as above: $(s_k{)}_{k\in \mathbb{N}}=\sum _{n=1}^k{a}_n$ is a Cauchy sequence, so for each $ϵ>0$ there is an $N\in \mathbb{N}$, such that for all $k\ge N$ there is

$|s_{2k-1}-s_{k-1}|\sum _{n=k}^{2k-1}{a}_n=a_k+a_{k+1}+\dots +a_{2k-2}+a_{2k-1}<ϵ$

Hence, for all $k\ge N$:

$\begin{array}{cc}\frac{1}{2}\cdot (2k-1)a_{2k-1}& \le \frac{1}{2}\cdot 2k{a}_{2k-1}\\ & =k{a}_{2k-1}\\ & =a_{2k-1}+a_{2k-1}+\dots +a_{2k-1}+a_{2k-1}\\ & \downarrow (a_n)\text{ decreases monotonically}\\ & \le a_k+a_{k+1}+\dots +a_{2k-2}+a_{2k-1}<ϵ\end{array}$

So we know that $(\frac{1}{2}\cdot (2k-1)a_{2k-1})$ and hence also $((2k-1)a_{2k-1})$ are null sequences.

The case distinction was actually only necessary since for even numbers $K=2k$, we had to sum over $k=\frac{K}{2}$ and for odd numbers $K=2k-1$ over $k=\frac{K+1}{2}$ elements. Since both for even $(2k{a}_{2k})$ and odd indices $((2k-1)a_{2k-1})$ , the elements tend to zero as $k\to \mathrm{\infty }$, the sequence $(n{a}_n)$ must also tend to zero, i.e. it is a null sequence.

Cauchy criterion: an application

Exercise (Alternating harmonic series)

Using the Cauchy criterion, prove that the alternating harmonic series $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{1}{k}$ converges. (note: the harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ does not converge!)

Solution (Alternating harmonic series)

Our aim is to show that the sequence of partial sums $s_n=\sum _{k=1}^n(-1{)}^{k+1}\frac{1}{k}$ is a Cauchy sequence. That means, we have to put a bound on $\begin{array}{cc}& |s_n-s_{m-1}|=|\sum _{k=m}^n(-1{)}^{k+1}\frac{1}{k}|\\ & =|(-1{)}^{m+1}\frac{1}{m}+(-1{)}^{m+2}\frac{1}{m+1}+(-1{)}^{m+3}\frac{1}{m+2}+(-1{)}^{m+4}\frac{1}{m+3}+(-1{)}^{m+5}\frac{1}{m+4}+\dots |\\ & =\underset{=1}{\underbrace{|(-1{)}^{m+1}|}}|\stackrel{\ge 0}{\overbrace{\frac{1}{m}\underset{\le 0}{\underbrace{-(\frac{1}{m+1}-\frac{1}{m+2})}}\underset{\le 0}{\underbrace{-(\frac{1}{m+3}-\frac{1}{m+4})}}\underset{\le 0}{\underbrace{-\dots }}}}|\\ & \le \left|\frac{1}{m}\right|\\ & =\frac{1}{m}\end{array}$

The sequence $(\frac{1}{m}{)}_{m\in \mathbb{N}}$ is definitely null. So for each $ϵ>0$ we can choose an $N\in \mathbb{N}$, such that $|s_n-s_{m-1}|=|\sum _{k=m}^n(-1{)}^{k+1}\frac{1}{k}|\le |\frac{1}{m}|<ϵ$ for all $n\ge m\ge N$. So $(s_n{)}_{n\in \mathbb{N}}$ is a Cauchy sequence implying that $\underset{n\to \mathrm{\infty }}{lim}{s}_n=\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{1}{k}$ converges, which was to be shown.

Root and ratio test: estimate of errors

In some cases, the infinite sum $s=\sum _{k=1}^{\mathrm{\infty }}{a}_k$ cannot be computed explicitly. We can try to approximate it by $s_n=\sum _{k=1}^n{a}_k$. The following exercises are devoted to giving upper bounds on the approximation error $|s-s_n|$:

Exercise (Error estimate for the root test)

Let $(a_n{)}_{n\in \mathbb{N}}$ be a sequence and $q<1$. Further, $\sqrt[n]{|a_n|}\le q$ for all $n\ge N$. Then, the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k=s$ converges absolutely by the root test. Show that it can be approximated by $s_n=\sum _{k=1}^n{a}_k$ with $n\ge N-1$ up to a maximal error of

$|s-s_n|\le \frac{q^{n+1}}{1-q}$

Solution (Error estimate for the root test)

The idea is that the root test bounds $\sum _k|a_k|$ by a geometric series $\sum _k{q}^k$ and $\sum _{k=n+1}^{\mathrm{\infty }}{q}^k=\frac{q^{n+1}}{1-q}$ This bounding can be explicitly shown by using the root test criterion:

$\sqrt[k]{|a_k|}\le q⟺|a_k|\le q^k$

So for $n\ge N-1⟺n+1\ge N$, we can sum up:

$\begin{array}{cc}|s-s_n|& =\left|\sum _{n+1}^{\mathrm{\infty }}{a}_k\right|\\ & \downarrow \text{generalized triangle inequality}\\ & \le \sum _{k=n+1}^{\mathrm{\infty }}|a_k|\\ & \le \sum _{k=n+1}^{\mathrm{\infty }}{q}^k\\ & \downarrow \text{index shift}\\ & =\sum _{k=0}^{\mathrm{\infty }}{q}^{n+1+k}\\ & =\sum _{k=0}^{\mathrm{\infty }}{q}^{n+1}\cdot q^k\\ & \downarrow \text{factorization rule}\\ & =q^{n+1}\cdot \sum _{k=0}^{\mathrm{\infty }}{q}^k\\ & \downarrow \text{geometric series}\\ & =q^{n+1}\cdot \frac{1}{1-q}=\frac{q^{n+1}}{1-q}\end{array}$

Exercise (Error estimate for the ratio test)

Let $(a_n{)}_{n\in \mathbb{N}}$ be a sequence and $q<1$. Further, let $a_n\ne 0$ and $\left|\frac{a_{n+1}}{a_n}\right|\le q$ for all $n\ge N$. By the ratio test, the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k=s$ converges absolutely. Show that it can be approximated by $s_n=\sum _{k=1}^n{a}_k$ with $n\ge N-1$ up to a maximal error of

$|s-s_n|\le \frac{|a_{n+1}|}{1-q}$

Solution (Error estimate for the ratio test)

Again, we bound the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ by a geometric series. But this time, the factor of $q^{n+1}$ is replaced by $|a_{n+1}|$. By assumption:

$\left|\frac{a_{k+1}}{a_k}\right|\le q⟺|a_{k+1}|\le q\cdot |a_k|$

For any $n\ge N-1⟺n+1\ge N$, we can conclude by iteration:

$\begin{array}{cc}& |a_{n+2}|\le q\cdot |a_{n+1}|\\ ⟹& |a_{n+3}|\le q\cdot |a_{n+2}|\le q^2\cdot |a_{n+1}|\\ ⟹& |a_{n+4}|\le q^3\cdot |a_{n+1}|\\ & \downarrow (k-3)\text{-fold iteration of the inequality}\\ ⟹& |a_{n+1+k}|\le q^k\cdot |a_{n+1}|\end{array}$

Again, we sum up the geometric series bounds:

$\begin{array}{cc}|s-s_n|& =\left|\sum _{n+1}^{\mathrm{\infty }}{a}_k\right|\\ & \downarrow \text{generalized triangle inequality}\\ & \le \sum _{k=n+1}^{\mathrm{\infty }}|a_k|\\ & \downarrow \text{index shift}\\ & =\sum _{k=0}^{\mathrm{\infty }}|a_{n+1+k}|\\ & \le \sum _{k=0}^{\mathrm{\infty }}{q}^k\cdot |a_{n+1}|\\ & \downarrow \text{factorization rule}\\ & =|a_{n+1}|\cdot \sum _{k=0}^{\mathrm{\infty }}{q}^k\\ & \downarrow \text{geometric series}\\ & =|a_{n+1}|\cdot \frac{1}{1-q}=\frac{|a_{n+1}|}{1-q}\end{array}$

Exercise (showing that a sequence is null)

1. Let $(a_n{)}_{n\in \mathbb{N}}$ be a sequence and $0<q<1$. Further, let $a_n\ne 0$ and $\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=q$ or $\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|a_n|}=q$. Show that we then have a null sequence $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$.
2. Conclude that $\underset{n\to \mathrm{\infty }}{lim}(\genfrac{}{}{0ex}{}{n}{k})q^n=0$ for $0<q<1$ and $k\in {\mathbb{N}}_0$.

Solution (showing that a sequence is null)

1. Both $\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=q<1$ (ratio test) and $\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|a_n|}=q<1$ (root test) imply that the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges. But by means of the term test , the sequence $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$ has to be a null sequence.
2. We specifically consider the sequence $a_n=(\genfrac{}{}{0ex}{}{n}{k})q^n$ . It satisfies the ratio test criterion:

$\begin{array}{cc}\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|& =\underset{n\to \mathrm{\infty }}{lim}\frac{(\genfrac{}{}{0ex}{}{n+1}{k})q^{n+1}}{(\genfrac{}{}{0ex}{}{n}{k})q^n}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{(n+1)n(n-1)\cdot \dots \cdot ((n+1)-k+1)}{k!}\cdot q\cdot q^n}{\frac{n(n-1)\cdot \dots \cdot (n-k+1)}{k!}\cdot q^n}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{(n+1)n(n-1)\cdot \dots \cdot (n-k+2)}{n(n-1)\cdot \dots \cdot (n-k+2)(n-k+1)}\cdot q\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{n+1}{n-k+1}\cdot q\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{1+\stackrel{\to 0}{\overbrace{\frac{1}{n}}}}{1-\underset{\to 0}{\underbrace{\frac{k}{n}}}+\underset{\to 0}{\underbrace{\frac{1}{n}}}}\cdot q\\ & =q<1\end{array}$

So the series $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges and by part 1 of the exercise, $(a_n{)}_{n\in \mathbb{N}}$ is a null sequence, i.e. $\underset{n\to \mathrm{\infty }}{lim}(\genfrac{}{}{0ex}{}{n}{k})q^n=0$.

Alternating series test with error bounds

Exercise (Alternating series test with error bounds)

First, show that the series

$\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{2k+1}\frac{k}{k+2}=:\sum _{k=1}^{\mathrm{\infty }}{a}_k=:S$

converges. Then, determine an index $n_0\in \mathbb{N}$, such that the partial sum $S_n=\sum _{k=1}^n\frac{(-1{)}^{k+1}}{2k+1}\frac{k}{k+2}$ for $n\ge n_0$ approximates $S$ up to a precision of $|S-S_N|\frac{1}{100}$ .

Solution (Alternating series test with error bounds)

Proof step: The series converges

For $b_k=|a_k|=\frac{k}{(2k+1)(k+2)}$ we can estimate

$\begin{array}{cc}\frac{b_{k+1}}{b_k}& =\frac{\frac{k+1}{(2k+3)(k+3)}}{\frac{k}{(2k+1)(k+2)}}\\ & =\frac{(k+1)(2k+1)(k+2)}{k(2k+3)(k+3)}\\ & =\frac{2k^3+7k^2+7k+2}{2k^3+9k^2+9k}\\ & =\frac{2k^3+7k^2+7k+2}{2k^3+7k^2+7k+2k^2+2k}\\ & =\frac{2k^3+7k^2+7k+2}{2k^3+7k^2+7k+2(k^2+k)}\\ & \downarrow k^2+k\ge 1\\ & \le \frac{2k^3+7k^2+7k+2}{2k^3+7k^2+7k+2}\\ & =1\end{array}$

So $(b_k{)}_{k\in \mathbb{N}}$ is monotonically decreasing. Further

$\begin{array}{cc}\underset{k\to \mathrm{\infty }}{lim}{b}_k& =\underset{k\to \mathrm{\infty }}{lim}\frac{k}{(2k+1)(k+2)}\\ & =\underset{k\to \mathrm{\infty }}{lim}\frac{k}{2k^2+5k+2}\\ & =\underset{k\to \mathrm{\infty }}{lim}\frac{\frac{1}{k}}{2+\frac{5}{k}+\frac{2}{k^2}}\\ & \downarrow \text{calculus for sequences}\\ & \le \frac{0}{2+0+0}\\ & =0\end{array}$

So $(b_k)$ is even a null sequence. By means of the alternating series test, the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ hence converges.

Proof step: Finding $n_0$

For the alternating series test, there is an error bound:

$|\sum _{k=1}^{\mathrm{\infty }}{b}_k-\sum _{k=1}^n{b}_k|<b_{n+1}$

With $b_{n+1}=\frac{n+1}{(2n+3)(n+3)}$. It saves a lot of calculation work to just give a coarse upper bound for this expression:

$b_{n+1}=\frac{n+1}{(2n+3)(n+3)}\le \frac{n+3}{(2n+3)(n+3)}=\frac{1}{2n+3}$

Now, if $\frac{1}{2n+3}<\frac{1}{100}$, then there is also $b_{n+1}<\frac{1}{100}$. Hence,

$\begin{array}{cc}\frac{1}{2n+3}<\frac{1}{100}& ⟺2n+3>100\\ & ⟺2n>97\\ & ⟺n>49\in \mathbb{N}\end{array}$

So $n+1=49$ (and any bigger $n$) leads to a precise enough approximation. For $n_0=n=48$ the approximation precision is therefore better than $|S-S_N|\le \frac{1}{100}$ .

Cauchy condensation test

Exercise (A double logarithm)

Determine, for which $\alpha >0$ the following series converges:

$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\ln (k)\ln (\ln (k){)}^{\alpha }}$

Solution (A double logarithm)

The Cauchy condensation test is useful to resolve the double logarithm: if in $\frac{1}{\ln (\ln (n))}$ we only consider the elements $n=2^k$ (i.e. 2, 4, 8, 16, ...), we get a single logarithm in $k$ instead of a double logarithm: $\frac{1}{\ln (\ln (n))}=\frac{1}{\ln (\ln (2^k))}=\frac{1}{\ln (k\ln 2)}$. As in our sequence, $a_k=\frac{1}{k\ln (k)\ln (\ln (k){)}^{\alpha }}$ is monotonically decreasing, the Cauchy condensation test yields that it converges if and only if the following series converges:

$\begin{array}{cc}\sum _{k=1}^{\mathrm{\infty }}{2}^k\frac{1}{2^k\ln (2^k)\ln (\ln (2^k){)}^{\alpha }}& =\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\ln (2^k)\ln (\ln (2^k){)}^{\alpha }}\\ & \downarrow \ln (x^y)=y\ln (x)\\ & =\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\ln (2)\ln (k\ln (2){)}^{\alpha }}\\ & \downarrow \ln (xy)=\ln (x)+\ln (y)\\ & =\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\ln (2)(\ln (k)+\ln (\ln (2)){)}^{\alpha }}\end{array}$

In the article "Cauchy condensation test", there is an exercise proving that $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n(\ln (n){)}^{\alpha }}$ converges for $\alpha >1$ and diverges for $0<\alpha \le 1$. This can be used for a direct comparison:

Case 1: $\alpha >1$

Here,

$\frac{1}{k\ln (2)(\ln (k)+\ln (\ln (2)){)}^{\alpha }}\le \frac{1}{k\ln (2)\ln (k{)}^{\alpha }}=\frac{1}{\ln (2)}\frac{1}{k\ln (k{)}^{\alpha }}$

and $\frac{1}{\ln (2)}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\ln (k{)}^{\alpha }}<\mathrm{\infty }$. So by direct comparison, our series converges for all $\alpha >1$.

Case 2: $0<\alpha \le 1$

Here,

$\frac{1}{k\ln (2)(\ln (k)+\ln (\ln (2)){)}^{\alpha }}\ge \frac{1}{k\ln (2)(\ln (k)+\ln (k){)}^{\alpha }}=\frac{1}{k\ln (2)(2\ln (k){)}^{\alpha }}=\frac{1}{k\ln (2)2^{\alpha }\ln (k{)}^{\alpha }}=\frac{1}{2^{\alpha }\ln (2)}\frac{1}{k\ln (k{)}^{\alpha }}$

and $\frac{1}{2^{\alpha }\ln (2)}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k\ln (k{)}^{\alpha }}$ diverges. So by direct comparison, the series diverges for all $0<\alpha \le 1$.

Further convergence criteria

Exercise (Series with products)

Let $(a_k{)}_{k\in \mathbb{N}}$ and $(b_k{)}_{k\in \mathbb{N}}$ be two real sequences. Prove:

1. Whenever the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges absolutely and $(b_k{)}_{k\in \mathbb{N}}$ is bounded, then also $\sum _{k=1}^{\mathrm{\infty }}{a}_k{b}_k$ converges absolutely.
2. There are pairs of a series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ , which converges and a sequence $(b_k{)}_{k\in \mathbb{N}}$ which is bounded, such that $\sum _{k=1}^{\mathrm{\infty }}{a}_k{b}_k$ does not converge. So the above statement does only hold true for absolute convergence, but not for convergence in general.

Solution (Series with products)

Part 1:

Way 1: Boundedness of partial sums

Since $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges absolutely, the sequence of partial sums ${(\sum _{k=1}^n|a_k|)}_{n\in \mathbb{N}}$ must be bounded. Further, $(b_k{)}_{k\in \mathbb{N}}$ is bounded. That means, there is an upper bound $S>0$ with $|b_k|\le S$ for all $k\in \mathbb{N}$ and therefore

$\sum _{k=1}^n|a_k{b}_k|\le \sum _{k=1}^n|a_k|\cdot S=S\cdot \sum _{k=1}^n|a_k|$

But now, ${(\sum _{k=1}^n|a_k|)}_{n\in \mathbb{N}}$ is bounded, so ${(S\cdot \sum _{k=1}^n|a_k|)}_{n\in \mathbb{N}}$ is bounded as well. The sequence ${(\sum _{k=1}^n|a_k{b}_k|)}_{n\in \mathbb{N}}$ is smaller than ${(S\cdot \sum _{k=1}^n|a_k|)}_{n\in \mathbb{N}}$ and therefore also bounded, i.e. $\sum _{k=1}^{\mathrm{\infty }}{a}_k{b}_k$ converges absolutely.

Way 2: Direct comparison test

As above, $(b_k{)}_{k\in \mathbb{N}}$ has an upper bound $S>0$ , meaning that $|b_k|\le S$ for all $k\in \mathbb{N}$. Hence,

$|a_k{b}_k|=|a_k|\cdot |b_k|\le S\cdot |a_k|$

Now, the series $\sum _{k=1}^{\mathrm{\infty }}|a_k|$ converges, as well as $S\cdot \sum _{k=1}^{\mathrm{\infty }}|a_k|$ . We use the direct comparison test: $\sum _{k=1}^{\mathrm{\infty }}|a_k{b}_k|$ is smaller than $S\cdot \sum _{k=1}^{\mathrm{\infty }}|a_k|$ and hence converges.

Part 2: We know that the harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ diverges, but can be made convergent by putting a minus sign in front of every second element $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k}{k}<\mathrm{\infty }$ . The idea is now to "reverse" this process: we take the convergent alternating series $a_k=\frac{(-1{)}^k}{k}$ and make it again divergent, turning every second element positive:

$\underset{=a_k}{\underbrace{\frac{(-1{)}^k}{k}}}\underset{=b_k}{\underbrace{(-1{)}^k}}=(-1{)}^{2k}\frac{1}{k}=\frac{1}{k}$

The "reversing sequence" $(b_k{)}_{k\in \mathbb{N}}$ is bounded, as $|b_k|=|(-1{)}^k|=1\le 1=S$. So $\sum _{k=1}^{\mathrm{\infty }}{a}_k=\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k}{k}$ converges and $(b_k{)}_{k\in \mathbb{N}}=((-1{)}^k{)}_{k\in \mathbb{N}}$ is bounded, but $\sum _{k=1}^{\mathrm{\infty }}{a}_k{b}_k=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ diverges as a harmonic series.

Note: other counter-examples are also possible.

Exercise (Raabe criterion)

1. Let $(a_n{)}_{n\in \mathbb{N}}$ and $(b_n{)}_{n\in \mathbb{N}}$ be two real sequences. Prove:

   If for almost all $n\in \mathbb{N}$ , there is $a_n\ne 0$ (all except for a finite number of elements). Then, if

   • $\left|\frac{a_{n+1}}{a_n}\right|\le 1-\frac{c}{n+1}$ for some $c>1$, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely.
   • $\left|\frac{a_{n+1}}{a_n}\right|\ge 1-\frac{1}{n+1}$ then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges.
   • Use this so-called Raabe criterion, to prove that the following series for any $s>0$ :
   $\sum _{n=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{s}{n})$

Solution (Raabe criterion)

Part 1:

• For convergence, we equivalently transform

$\begin{array}{cc}\left|\frac{a_{n+1}}{a_n}\right|\le 1-\frac{c}{n+1}& \stackrel{\cdot (n+1)|a_n|}{⟺}(n+1)\left|a_{n+1}\right|\le (n+1)|a_n|-c|a_n|\\ & \downarrow \text{ add }(c-1)|a_n|\text{ and subtract }(n+1)|a_{n+1}|\text{ on both sides}\\ & ⟺(c-1)\left|a_n\right|\le n|a_n|-(n+1)|a_{n+1}|\\ & \downarrow \text{ divide both sides by }c-1>0\\ & ⟺\left|a_n\right|\le \frac{1}{c-1}\cdot (n|a_n|-(n+1)|a_{n+1}|)\end{array}$

This works for almost all $n\in \mathbb{N}$ . So we can find an $n_0\in \mathbb{N}$, such that the re-formulation works for all $n\ge n_0$ . We take the sum of both sides starting from $n_0$ up to some $N>n_0$:

$\begin{array}{cc}\sum _{n=n_0}^N\left|a_n\right|& \le \frac{1}{c-1}\cdot \sum _{n=n_0}^N(n|a_n|-(n+1)|a_{n+1}|)\\ & \downarrow \text{ telescoping sum}\\ & =\frac{1}{c-1}\cdot (n_0|a_{n_0}|-(N+1)|a_{N+1}|)\\ & \le \frac{1}{c-1}\cdot n_0|a_{n_0}|\end{array}$

Therefore, the sequence of partial sums ${\left(\sum _{n=n_0}^N\left|a_n\right|\right)}_{N\in \mathbb{N}}$ is bounded. That means, the series $\sum _{n=n_0}^{\mathrm{\infty }}\left|a_n\right|$ converges absolutely. Adding a finite number of elements, we get that also $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges absolutely.

• In the case, where divergence should hold, we can equivalently transform for all $n\ge n_0$ :

$\begin{array}{cc}\left|\frac{a_{n+1}}{a_n}\right|\ge 1-\frac{1}{n+1}& \stackrel{\cdot (n+1)|a_n|}{⟺}(n+1)\left|a_{n+1}\right|\ge (n+1)|a_n|-|a_n|\\ & ⟺(n+1)\left|a_{n+1}\right|\ge n|a_n|\\ & \downarrow \text{ subsequently apply the inequality}\\ & ⟺(n+1)\left|a_{n+1}\right|\ge n|a_n|\ge (n-1)|a_{n-1}|\ge \dots \ge n_0|a_{n_0}|\end{array}$

This is equivalent to

$\left|a_{n+1}\right|\ge n_0|a_{n_0}|\cdot \frac{1}{n+1}$

The harmonic series $\sum _{n=n_0}^{\mathrm{\infty }}\underset{\text{fest}}{\underbrace{n_0\left|a_{n_0}\right|}}\cdot \frac{1}{n+1}$ diverges. By direct comparison, this implies that also $\sum _{n=n_0}^{\mathrm{\infty }}{a}_n$ diverges. Adding a finite number of elements, also $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges, which was to be shown.

Part 2: Here, $a_n=(\genfrac{}{}{0ex}{}{s}{n})$, which means

$\begin{array}{cc}\left|\frac{a_{n+1}}{a_n}\right|& =\frac{(\genfrac{}{}{0ex}{}{s}{n+1})}{(\genfrac{}{}{0ex}{}{s}{n})}\\ & =\left|\frac{\frac{s(s-1)\cdot \dots \cdot (s-n+1)(\stackrel{=s-n}{\overbrace{s-(n+1)+1}})\cdot n!}{(n+1)!}}{\frac{s(s-1)\cdot \dots \cdot (s-n+1)}{n!}}\right|\\ & =\left|\frac{s(s-1)\cdot \dots \cdot (s-n+1)(s-n)n!}{s(s-1)\cdot \dots \cdot (s-n+1)\cdot (n+1)n!}\right|\\ & =\frac{|s-n|}{n+1}\\ & \downarrow \text{ for }n>s\\ & =\frac{n-s}{n+1}\\ & =\frac{n+1-s-1}{n+1}\\ & =1-\frac{s+1}{n+1}\\ & \le 1-\frac{s+1}{n+1}\end{array}$

So $c=s+1>1$ is a suitable constant to use the Raabe criterion and imply absolute convergence of the series $\sum _{n=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{s}{n})$.