Alternating series test – "Math for Non-Geeks"

An introductory video to the alternating series test. (German) (YouTube-Video , channel: Quatematik)

The alternating series criterion serves to prove convergence of an alternating series, i.e. a series where the pre-signs alternately change from positive to negative, like $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ or $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k{b}_k$ (with all $b_k$ being positive). Series of this kind can be convergent, but not absolutely convergent. In those cases, criteria for absolute convergence will fail, but the alternating series criterion may be successful.

The alternating series criterion goes back to Gottfried Wilhelm Leibniz and was published in 1682.

Introductory example: Convergence of the alternating harmonic series

Series treated by the alternating series criterion will often converge, but not converge absolutely. Perhaps, the most prominent example for such a converging, but not absolutely convergent series is the alternating harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k}$. Convergence of it can be shown by making sure that the sequence of partial sums $(S_n{)}_{n\in \mathbb{N}}={\left(\sum _{k=1}^n\frac{(-1{)}^{k+1}}{k}\right)}_{n\in \mathbb{N}}$ converges. For $n=1,2,3,4,5,6,7,8$ , those partial sums are

$\begin{array}{cccc}{S}_1& =\sum _{k=1}^1\frac{(-1{)}^{k+1}}{k}=1,& S_2& =\sum _{k=1}^2\frac{(-1{)}^{k+1}}{k}=\frac{1}{2}=0.5,\\ S_3& =\sum _{k=1}^3\frac{(-1{)}^{k+1}}{k}=\frac{5}{6}=0.8\bar{3},& S_4& =\sum _{k=1}^4\frac{(-1{)}^{k+1}}{k}=\frac{7}{12}=0.58\bar{3},\\ S_5& =\sum _{k=1}^5\frac{(-1{)}^{k+1}}{k}=\frac{47}{60}=0.78\bar{3},& S_6& =\sum _{k=1}^6\frac{(-1{)}^{k+1}}{k}=\frac{37}{60}=0.61\bar{6},\\ S_7& =\sum _{k=1}^7\frac{(-1{)}^{k+1}}{k}=\frac{329}{420}=0.759,& S_8& =\sum _{k=1}^8\frac{(-1{)}^{k+1}}{k}=\frac{533}{840}=\mathrm{0.634.}\end{array}$

Those partial sums tend to make jumps of ever smaller getting distance. Those partial sums with odd indices, ($S_{2n-1}$) seem to be monotonically decreasing and those with even indices $S_{2n}$ seem to be increasing. A simple calculation can mathematically verify this assertion: For all $n\in \mathbb{N}$ there is

$\begin{array}{cc}{S}_{2n+1}-S_{2n-1}& =\sum _{k=1}^{2n+1}\frac{(-1{)}^{k+1}}{k}-\sum _{k=1}^{2n-1}\frac{(-1{)}^{k+1}}{k}\\ & =(1-\frac{1}{2}+\frac{1}{3}-\dots +\frac{(-1{)}^{2n}}{2n-1}+\frac{(-1{)}^{2n+1}}{2n}+\frac{(-1{)}^{2n+2}}{2n+1})\\ & -(1-\frac{1}{2}+\frac{1}{3}-\dots +\frac{(-1{)}^{2n}}{2n-1})\\ & =\frac{(-1{)}^{2n+1}}{2n}+\frac{(-1{)}^{2n+2}}{2n+1}\\ & =\frac{1}{2n+1}-\frac{1}{2n}\\ & \downarrow 2n+1\ge 2n\Rightarrow \frac{1}{2n+1}\le \frac{1}{2n}\\ & \le 0,\end{array}$

i.e. $S_{2n+1}\le S_{2n-1}$ (monotonically decreasing). Analogously, $S_{2n+2}-S_{2n}=\frac{(-1{)}^{2n+3}}{2n+2}+\frac{(-1{)}^{2n+2}}{2n+1}=-\frac{1}{2n+2}+\frac{1}{2n+1}\ge 0$, so $S_{2n+2}\ge S_{2n}$ (monotonically increasing).

If we could now show that $(S_{2n-1})$ is bounded from below and $(S_{2n})$is bounded from above, then both sequences would converge by the monotonicity criterion. Luckily, this is exactly the case: all odd partial sums are bounded from below by any even partial sum and all even partial sums are bounded from above by any odd partial sum: For all $n\in \mathbb{N}$

$S_{2n-1}-S_{2n}=-\frac{(-1{)}^{2n+1}}{2n}=\frac{(-1{)}^{2n+2}}{2n}=\frac{1}{2n}\ge 0,$

so $S_{2n-1}\ge S_{2n}$ and $S_{2n}\le S_{2n-1}$. We therefore have the bounds $S_{2n-1}\ge S_{2n}\ge S_2=\frac{1}{2}$ and $S_{2n}\le S_{2n-1}\le S_1=1$. Hence, $(S_{2n-1})$ is bounded from below by $\frac{1}{2}$ and $(S_{2n})$ is bounded from above by $1$ .

The monotonicity criterion now implies that both the subsequences of partial sums $(S_{2n-1})$ and $(S_{2n})$ are convergent.

In order to get convergence of the series, we need to show that the sequence of partial sums $(S_n)$ converges. This is for sure the case, if both the odd and the even subsequence $(S_{2n-1})$ and $(S_{2n})$ converge to the same limit.

How can this be shown? First, let us assign a name too the limits: $\underset{n\to \mathrm{\infty }}{lim}{S}_{2n-1}=S$ and $\underset{n\to \mathrm{\infty }}{lim}{S}_{2n}=S^{\prime }$. The statement we want to show can then mathematically be expressed as $S=S^{\prime }$. We show this by subtracting both limits from each other, which is equivalent to taking the limit of the sequence difference:

$S-S^{\prime }=\underset{n\to \mathrm{\infty }}{lim}{S}_{2n-1}-\underset{n\to \mathrm{\infty }}{lim}{S}_{2n}=\underset{n\to \mathrm{\infty }}{lim}(S_{2n-1}-S_{2n}).$

Above, we showed $S_{2n-1}-S_{2n}=\frac{1}{2n}$ which is a null sequence:

$\underset{n\to \mathrm{\infty }}{lim}(S_{2n-1}-S_{2n})=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2n}=0.$

Hence, $S-S^{\prime }=0$ which means $S=S^{\prime }$.

From this, we can imply that the sequence of partial sums $(S_n)$ converges to $S$ . Mathematically, we need to stay closer than any $ϵ$ to the limit value after surpassing some sequence element number $N$ for the corresponding $ϵ$

$\mathrm{\forall }ϵ>0\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }n\ge N:|S_n-S|<ϵ.$

For a fixed $ϵ$, both the odd and the even partial sum sequences have a suitable $N$, which we name by $(2N_1-1)$ (odd) and $2N_2$(even):

$\begin{array}{cc}& \mathrm{\forall }ϵ>0\mathrm{\exists }{N}_1\in \mathbb{N}\mathrm{\forall }n\ge N_1:|S_{2n-1}-S|<ϵ\\ & \mathrm{\forall }ϵ>0\mathrm{\exists }{N}_2\in \mathbb{N}\mathrm{\forall }n\ge N_2:|S_{2n}-S|<ϵ,\end{array}$

After reaching the greater of these two numbers $N=max\{(2N_1-1),2N_2\}$, both sequences stay closer than $ϵ$ to the limit value and

$\mathrm{\forall }ϵ>0\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }n\ge N:|S_n-S|<ϵ.$

Generalizing the proof idea / alternating series test

Now we consider any alternating series. Can we use the same proof as for the alternating harmonic series to show that our general alternating series converges? The answer will depend on the properties of the general alternating series. We used the following properties from the alternating harmonics series:

• The sequence of coefficients $(b_k)=\left(\frac{1}{k}\right)$ without the alternating sign is monotonically decreasing. This gave us monotonicity and boundedness of the two partial sums $(S_{2n-1})$ and $(S_{2n})$ , so we could show that they converge. Without the monotonicity, this may not be the case.
• Further, we used that $(b_k)=\left(\frac{1}{k}\right)$ is a null sequence. This was needed to show that both $(S_{2n-1})$ and $(S_{2n})$ had the same limit, so $(S_n)$ converges to that limit. If $(b_k)=\left(\frac{1}{k}\right)$ converged to a constant $B$, then $(S_n)$ would in the end tend to "jump" up and down by an amount of $B$ and the limits of $(S_{2n-1})$ and $(S_{2n})$ may differ by $B$, so they are not equal.

No further properties of the alternating harmonic series have been used for the proof. So we may use the above proof steps to show convergence of a general alternating series:

Theorem (Alternating series test)

Let $(b_k{)}_{k\in \mathbb{N}}$ be a non-negative, monotonically decreasing null sequence of real numbers, i.e. $\underset{k\to \mathrm{\infty }}{lim}{b}_k=0$. Then, the alternating series $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ converges.

The proof uses the same steps as the convergence proof for the alternating harmonic series above.

Proof (Alternating series test)

We need to show that the sequence of partial sums $(S_n)=(\sum _{k=1}^n(-1{)}^{k+1}{b}_k)$ converges.

Step 1: The odd subsequence $(S_{2n-1})$ is monotonically decreasing and the even subsequence $(S_{2n})$ is monotonically increasing, as for any $n\in \mathbb{N}$ there is

$\begin{array}{cc}{S}_{2n+1}-S_{2n-1}& =\sum _{k=1}^{2n+1}(-1{)}^{k+1}{b}_k-\sum _{k=1}^{2n-1}(-1{)}^{k+1}{b}_k\\ & =(b_1-b_2+b_3-\dots +(-1{)}^{2n}{b}_{2n-1}\\ & +(-1{)}^{2n+1}{b}_{2n}+(-1{)}^{2n+2}{b}_{2n+1})\\ & -(b_1-b_2+b_3-\dots +(-1{)}^{2n}{b}_{2n-1})\\ & =(-1{)}^{2n+2}{b}_{2n+1}+(-1{)}^{2n+1}{b}_{2n}\\ & =b_{2n+1}-b_{2n}\\ & \downarrow b_{2n+1}\le b_{2n}\\ & \le 0,\end{array}$

and analogously $S_{2n+2}-S_{2n}=(-1{)}^{2n+3}{b}_{2n+2}+(-1{)}^{2n+2}{b}_{2n+1}=-b_{2n+2}+b_{2n+1}\ge 0$.

Step 2: $(S_{2n-1})$ is bounded from below and $(S_{2n})$ is bounded from above, since for $n\in \mathbb{N}$ there is

$S_{2n-1}-S_{2n}=(-1{)}^{2n+2}{b}_{2n}=b_{2n}\ge 0$

So $S_{2n-1}\ge S_{2n}\ge S_2=b_1-b_2$ and $S_{2n}\le S_{2n-1}\le S_1=b_1$

The monotonicity criterion yields convergence of both the partial sums $(S_{2n-1})$ and $(S_{2n})$.

Step 3: $(S_{2n-1})$ and $(S_{2n})$ converge to the same limit. Let $\underset{n\to \mathrm{\infty }}{lim}{S}_{2n-1}=S$ and $\underset{n\to \mathrm{\infty }}{lim}{S}_{2n}=S^{\prime }$. In step 2, we proved convergence of both sequences, so we can use the sum rule for limits of sequences:

$S-S^{\prime }=\underset{n\to \mathrm{\infty }}{lim}{S}_{2n-1}-\underset{n\to \mathrm{\infty }}{lim}{S}_{2n}=\underset{n\to \mathrm{\infty }}{lim}(S_{2n-1}-S_{2n})$

On the other hand,

$\underset{n\to \mathrm{\infty }}{lim}(S_{2n-1}-S_{2n})=\underset{n\to \mathrm{\infty }}{lim}{b}_{2n}=0,$

since both $(b_k)$ and $(b_{2n})$ are null sequences. So both limits are equal ($S=S^{\prime }$).

Step 4: $(S_n)$ also converges to $S$. Since $(S_{2n-1})$ and $(S_{2n})$ converge to $S$ , both approach $S$ up to any $ϵ$:

$\begin{array}{cc}& \mathrm{\forall }ϵ>0\mathrm{\exists }{N}_1\in \mathbb{N}\mathrm{\forall }n\ge N_1:|S_{2n-1}-S|<ϵ\\ & \mathrm{\forall }ϵ>0\mathrm{\exists }{N}_2\in \mathbb{N}\mathrm{\forall }n\ge N_2:|S_{2n}-S|<ϵ\end{array}$

We now take the greater number $N=max\{2N_1-1,2N_2\}$ and obtain that also $(S_n)$ approaches $S$ up to $ϵ$:

$\mathrm{\forall }ϵ>0\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }n\ge N:|S_n-S|<ϵ$

So the series $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ converges - and we are done with the proof.

Alternative proof

Alternatively, one may use the Cauchy criterion for proving the alternating series criterion.

Alternative proof (Alternating series test)

In order to apply the alternating series criterion, we need to show that (under the assumptions of the alternating series test), there is $\mathrm{\forall }ϵ>0\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }n\ge m\ge N:|\sum _{k=m}^n(-1{)}^{k+1}{b}_k|<ϵ.$

At first, let us take only odd $m$ and estimate the sum $\sum _{k=m}^n(-1{)}^{k+1}{b}_k$ . This sum is positive: $\begin{array}{cc}\sum _{k=m}^n(-1{)}^{k+1}{b}_k& =(-1{)}^{m+1}{b}_m+(-1{)}^{m+2}{b}_{m+1}+(-1{)}^{m+3}{b}_{m+2}+(-1{)}^{m+4}{b}_{m+3}+\dots \\ & \downarrow m\text{ odd}\Rightarrow m+1\text{ even}\Rightarrow m+2\text{ odd}\\ & =b_m-b_{m+1}+b_{m+2}-b_{m+3}+\dots +\{\begin{array}{cc}{b}_n& \text{ if }n\text{ odd,}\\ (-b_n)& \text{ if }n\text{ even}\end{array}\\ & \downarrow \text{ associativity}\\ & =(b_m-b_{m+1})+(b_{m+2}-b_{m+3})+\dots +\{\begin{array}{cc}{b}_n& \text{ if }n\text{ odd,}\\ (b_{n-1}-b_n)& \text{ if }n\text{ even}\end{array}\\ & \downarrow (b_k)\text{ monotonically decreasing }\Rightarrow b_k-b_{k+1}\ge 0\\ & =\underset{\ge 0}{\underbrace{(b_m-b_{m+1})}}+\underset{\ge 0}{\underbrace{(b_{m+2}-b_{m+3})}}+\dots +\{\begin{array}{cc}\underset{\ge 0}{\underbrace{b_n}}& \text{ if }n\text{ odd,}\\ \underset{\ge 0}{\underbrace{(b_{n-1}-b_n)}}& \text{ if }n\text{ even}\end{array}\\ & \ge 0.\end{array}$

But it is also bounded:

$\begin{array}{cc}\sum _{k=m}^n(-1{)}^{k+1}{b}_k& =(-1{)}^{m+1}{b}_m+(-1{)}^{m+2}{b}_{m+1}+(-1{)}^{m+3}{b}_{m+2}+(-1{)}^{m+4}{b}_{m+3}+\dots \\ & \downarrow m\text{ odd}\Rightarrow m+1\text{ even}\Rightarrow m+2\text{ odd }\\ & =b_m-b_{m+1}+b_{m+2}-b_{m+3}+\dots +\{\begin{array}{cc}{b}_n& \text{ if }n\text{ odd,}\\ (-b_n)& \text{ if }n\text{ even}\end{array}\\ & \downarrow \text{ associativity}\\ & =b_m-(b_{m+1}-(b_{m+2})-(b_{m+3}-b_{m+4})+\dots +\{\begin{array}{cc}-(b_{n-1}-b_n)& \text{ if }n\text{ odd,}\\ -b_n& \text{ if }n\text{ even}\end{array}\\ & \downarrow (b_k)\text{ monotonically decreasing }\Rightarrow b_k-b_{k+1}\ge 0\\ & =b_m\underset{\le 0}{\underbrace{-(b_{m+1}-b_{m+2})}}\underset{\le 0}{\underbrace{-(b_{m+3}-b_{m+4})}}+\dots +\{\begin{array}{cc}\underset{\le 0}{\underbrace{-(b_{n-1}-b_n)}}& \text{ if }n\text{ odd,}\\ \underset{\le 0}{\underbrace{-b_n}}& \text{ if }n\text{ even}\end{array}\\ & \le b_m.\end{array}$

So, for odd $m$, there is $|\sum _{k=m}^n(-1{)}^{k+1}{b}_k|=\sum _{k=m}^n(-1{)}^{k+1}{b}_k\le b_m=|b_m|$.

Analogously, for even $m$ , the sum is negative and bounded:

$\sum _{k=m}^n(-1{)}^{k+1}{b}_k\le 0\text{ and }\sum _{k=m}^n(-1{)}^{k+1}{b}_k\le -b_m,$

And the absolute value of the sum is also bounded by $|\sum _{k=m}^n(-1{)}^{k+1}{b}_k|\le |b_m|$ . The latter inequality holds for odd and even, i.e. for all $m\in \mathbb{N}$.

But now, $(b_k)$ was assumed to be a null sequence, so $\mathrm{\forall }ϵ>0\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }m\ge N:|b_m|<ϵ$. The inequality for the absolute value hence implies

$\mathrm{\forall }ϵ>0\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }n\ge m\ge N:|\sum _{k=m}^n(-1{)}^{k+1}{b}_k|<ϵ.$

And hence, the series $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ converges by the Cauchy criterion.

Application example

Example (Generalized alternating harmonic series)

The generalized alternating harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k^{\alpha }}$ converges for all $\alpha >0$ by the alternating series test, since

• $b_k=\frac{1}{k^{\alpha }}\ge 0$ for all $k\in \mathbb{N}$.
• $b_{k+1}=\frac{1}{(k+1{)}^{\alpha }}\stackrel{(k+1{)}^{\alpha }\ge k^{\alpha }}{\le }\frac{1}{k^{\alpha }}$ for all $k\in \mathbb{N}$, so $(b_k)$ is monotonically decreasing.
• $\underset{k\to \mathrm{\infty }}{lim}{b}_k=\underset{k\to \mathrm{\infty }}{lim}\frac{1}{k^{\alpha }}=0$, so $(b_k)$ is a null sequence.

Note that for $\alpha >1$, this series even converges absolutely, as $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}=\zeta (\alpha )<\mathrm{\infty }$.

Notes to the alternating series test

• Of course, we can also change the series signs from positive $\to$ negative $\to$ positive $\to \mathrm{...}$ to negative $\to$ positive $\to$ negative $\to \mathrm{...}$ and get a valid convergence criterion for series like $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k{b}_k$. The proof is the same, under an interchange of $(S_{2n-1})$ and $(S_{2n})$.
• One can also start from $k=0$, i.e. consider series like $\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{b}_k$ or $\sum _{k=0}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$. Any starting index $k$ is OK. The proof is just the same including an index shift.
• As above, the alternating series test does only lead convergence, but no absolute convergence. For instance, the alternating harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k}$ converges by the alternating series test. However, it does not converge absolutely.
• The alternating series test can never be used for implying divergence of a series. If a series fails to meet the criteria for the alternating series test, it can still converge. There is an example warning about this below

• We could also take $(b_k{)}_{k\in \mathbb{N}}$ to be a non-positive and monotonically increasing null sequence. I.e. it approaches $0$ from below. The proof works the same way. Especially, that means $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ converges, whenever $(b_k{)}_{k\in \mathbb{N}}$ is just any monotone null sequence.

A test problem

Exercise (Alternating series test)

Is the series $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{\sqrt{k}}{k+1}$ convergent?

How to get to the proof? (Alternating series test)

In order to apply the alternating series test, we need to show that the sequence $(b_k)=\left(\frac{\sqrt{k}}{k+1}\right)$ is a monotonically decreasing null sequence. That means, we need 2 proof steps:

1. We prove: $(b_k)$ is monotonically decreasing, i.e. $b_{k+1}\le b_k$. This can be shown by proving that the ratio between two elements is $\frac{b_{k+1}}{b_k}\le 1$ or their difference is $b_{k+1}-b_k\le 0$ . Differences of square roots are hard to handle, so we investigate the ratio $\frac{b_{k+1}}{b_k}\le 1$.
2. We prove: $(b_k)$ is a null sequence, i.e. $\underset{k\to \mathrm{\infty }}{lim}{b}_k=0$. We can prove this using limit theorems for sequences.

Proof (Alternating series test)

The sequence $(b_k)=\left(\frac{\sqrt{k}}{k+1}\right)$ is nonnegative ($b_k\ge 0$) and

Step 1: $\frac{b_{k+1}}{b_k}\le 1$

$\begin{array}{cc}\frac{b_{k+1}}{b_k}& =\frac{\frac{\sqrt{k+1}}{k+2}}{\frac{\sqrt{k}}{k+1}}\\ & =\frac{\sqrt{k+1}(k+1)}{\sqrt{k}(k+2)}\\ & =\frac{\sqrt{(k+1)(k+1{)}^2}}{\sqrt{k(k+2{)}^2}}\\ & =\frac{\sqrt{k^3+3k^2+3k+1}}{\sqrt{k^3+4k^2+4k}}\\ & =\sqrt{\frac{k^3+3k^2+3k+1}{k^3+4k^2+4k}}\end{array}$

Now, the square root defines a strictly monotonically increasing function. Hence, it is $\le 1$, whenever the number under the root is $\le 1$. We investigate the behaviour of the expression under the root in the limit $k\to \mathrm{\infty }$:

$\begin{array}{cc}\frac{k^3+3k^2+3k+1}{k^3+4k^2+4k}& =\frac{k^3+3k^2+3k+1}{k^3+3k^2+3k+(k^2+k)}\\ & \downarrow k^2+k\ge 1\\ & \le \frac{k^3+3k^2+3k+1}{k^3+3k^2+3k+1}\\ & =1\end{array}$

Hence, $\frac{b_{k+1}}{b_k}\le 1$, so $(b_k)$ is monotonically decreasing.

Step 2: $\underset{k\to \mathrm{\infty }}{lim}{b}_k=0$

$\begin{array}{cc}\underset{k\to \mathrm{\infty }}{lim}{b}_k& =\underset{k\to \mathrm{\infty }}{lim}\frac{\sqrt{k}}{k+1}\\ & \downarrow \text{ factorize the denominator }\\ & =\underset{k\to \mathrm{\infty }}{lim}\frac{\sqrt{k}}{\sqrt{k}(\sqrt{k}+\frac{1}{\sqrt{k}})}\\ & \downarrow \text{cancel }\sqrt{k}\\ & =\underset{k\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{k}+\frac{1}{\sqrt{k}}}\\ & \downarrow \underset{k\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{k}}=0\text{ use limit theorems}\\ & =0\end{array}$

So $(b_k)$ is a null sequence.

Additional question: Does this series converge absolutely?

No, $\sum _{k=1}^{\mathrm{\infty }}|(-1{)}^{k+1}\frac{\sqrt{k}}{k+1}|=\sum _{k=1}^{\mathrm{\infty }}\frac{\sqrt{k}}{k+1}$ diverges. For large $k$, the elements scale like $\frac{\sqrt{k}}{k+1}\propto \frac{1}{\sqrt{k}}$, which is even larger than a harmonic series $\frac{1}{k}$. More precisely,

$\begin{array}{cc}\frac{\sqrt{k}}{k+1}& \ge \frac{\sqrt{k}}{k+k}\\ & =\frac{\sqrt{k}}{2k}\\ & =\frac{1}{2\sqrt{k}}\\ & \ge \frac{1}{2k}\end{array}$

and $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{2k}$ is a divergent harmonic series.

So we have divergence by direct comparison.

What, if a condition is not fulfilled?

It is important to check that the 2 conditions for the alternating series test are fulfilled! There are alternating series, which do not meet one. The following examples will illustrate alternating series, where $(b_k)$ is either not converging to 0 (our example converges to 1) or not monotone. Both examples fail to be convergent (although they are alternating). The third example is an alternating series, which fails the alternating series test (as it is not monotone), but nevertheless converges. So the alternating series test does not identify all convergent alternating series.

Example (Example 1: Convergence to 0 is needed.)

We consider the series $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k\left(\frac{k+1}{k}\right)$ with $b_k=\frac{k+1}{k}$. Does it fulfil the requirements for the alternating series test? Let us check:

• $(b_k)$ is monotonically decreasing.

Exercise (Monotonicity)

Prove it.

Proof (Monotonicity)

We investigate the ratio $\frac{b_{k+1}}{b_k}=\frac{\frac{k+2}{k+1}}{\frac{k+1}{k}}=\frac{k(k+2)}{(k+1{)}^2}=\frac{k^2+2k}{k^2+2k+1}\le 1$ , so $b_{k+1}\le b_k$. Hence, $(b_k)$ is monotonically decreasing.

• $(b_k)$ converges to 1, so it is not a null sequence. We can easily see this by explicitly computing the limit: $\underset{k\to \mathrm{\infty }}{lim}{b}_k=\underset{k\to \mathrm{\infty }}{lim}\frac{k+1}{k}=\underset{k\to \mathrm{\infty }}{lim}\frac{1+\frac{1}{k}}{1}=\frac{1}{1}=1\ne 0$.

So the alternating series test can not be applied. And indeed, the series diverges, since for large $k$, it essentially behaves like $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^k\cdot 1$ . More precisely, the divergence can be proven by the term test: The sequence of elements $(a_k)=((-1{)}^k\frac{k+1}{k})$ can not be a null sequence, since the subsequence $(a_{2n})$ converges to 1. So the corresponding series diverges by the term test.

Example (Example 2: Monotonicity is needed.)

Next, we consider the series $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ with $b_k=\{\begin{array}{cc}\frac{1}{\sqrt{k}}& \text{ for odd }k,\\ \frac{1}{k}& \text{ for even }k\end{array}$. Will it converge? Let us try the alternating series test:

• $(b_k)$ is not monotonically decreasing. This can be easily seen considering the first series elements: $(1,\frac{1}{2},\frac{1}{\sqrt{3}},\frac{1}{4},\frac{1}{\sqrt{5}},\frac{1}{6},\frac{1}{\sqrt{7}},\frac{1}{8},\frac{1}{\sqrt{9}}=\frac{1}{3},\frac{1}{10},\dots )$. Clearly, $\frac{1}{2}<\frac{1}{\sqrt{3}},\frac{1}{4}<\frac{1}{\sqrt{5}},\frac{1}{6}<\frac{1}{\sqrt{7}},\frac{1}{8}<\frac{1}{\sqrt{9}}=\frac{1}{3}$ . More generally, for all $n\in \mathbb{N}$ there is $2n+1\le 4n^2$ (can be shown by induction) and hence $a_{2n}=\frac{1}{2n}=\frac{1}{\sqrt{4n^2}}\le \frac{1}{\sqrt{2n+1}}=a_{2n+1}$. So the sequence of elements is not monotonically decreasing, since otherwise we would have $a_{2n+1}\le a_{2n}$ . Only the subsequence of elements with even and with odd index would be monotonically decreasing on their own.
• However, $(b_k)$ is a null sequence.

Exercise (Null sequence)

Prove it.

Proof (Null sequence)

The subsequences of elements with only odd or only even index are both null sequences : $\underset{n\to \mathrm{\infty }}{lim}{b}_{2n-1}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{2n-1}}=0$ and $\underset{n\to \mathrm{\infty }}{lim}{b}_{2n}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2n}=0$. So $\mathrm{\forall }ϵ>0\mathrm{\exists }{N}_1\in \mathbb{N}\mathrm{\forall }n\ge N_1:|b_{2n-1}-0|=\frac{1}{\sqrt{2n-1}}<ϵ$ and $\mathrm{\forall }ϵ>0\mathrm{\exists }{N}_2\in \mathbb{N}\mathrm{\forall }n\ge N_2:|b_{2n}-0|=\frac{1}{2n}<ϵ$

(More precisely, some suitable $N$ numbers are provided by $N_1>\frac{1}{2{ϵ}^2}+\frac{1}{2}$ and $N_2>\frac{1}{2ϵ}$.) Now, we set $N=max\{2N_1-1,2N_2\}$. So if the index surpasses $N$ , it will also have surpassed $2N_1-1$ and $2N_2$, so both subsequences stay closer to 0 than $ϵ$, which means that the entire sequence will stay closer to 0 than $ϵ$:

$\mathrm{\forall }ϵ>0\mathrm{\exists }N\in \mathbb{N}\mathrm{\forall }k\ge N:|b_k-0|=b_k<ϵ.$ Therefore, $(b_k)$ converges to $0$.

Hence, the alternating series test does not apply. In fact, we can show that the sequence of partial sums $(s_{2n})=(\sum _{k=1}^{2n}(-1{)}^k{b}_k)$ is unbounded, so the series diverges. We use the estimate

$(k-1{)}^2\ge 0\Rightarrow k^2\ge 2k-1\Rightarrow k\ge \sqrt{2k-1}\Rightarrow \frac{1}{\sqrt{2k-1}}\ge \frac{1}{k}\Rightarrow \frac{1}{\sqrt{2k-1}}-\frac{1}{2k}\ge \frac{1}{2k}$

This implies

$\begin{array}{cc}{s}_{2n}& =\sum _{k=1}^{2n}(-1{)}^{k+1}{b}_k\\ & =(b_1-b_2)+(b_3-b_4)+\dots +(b_{2n-1}-b_{2n})\\ & =\sum _{k=1}^n(b_{2k-1}-b_{2k})\\ & =\sum _{k=1}^n(\frac{1}{\sqrt{2k-1}}-\frac{1}{2k})\\ & \downarrow \text{estimate above}\\ & \ge \sum _{k=1}^n\frac{1}{2k}\end{array}$

Since the harmonic series diverges, also $(s_{2n})$ and the entire series will diverge. Loosely speaking, the reason for the divergence is that the series corresponding to the positive and negative elements, $\frac{1}{\sqrt{k}}$ and $\frac{1}{k}$ diverge to $\mathrm{\infty }$ at different speeds. And their speed difference is so huge that the series of differences diverges, too.

Example (Example 3: A converging alternating series, which fails the alternating series test.)

Let us consider the series $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ with $b_k=\{\begin{array}{cc}\frac{1}{k^2}& \text{ for even }k,\\ 0& \text{ for odd }k\end{array}$. Now,

• $(b_k)$ is not monotonically decreasing, as for all $k\in \mathbb{N}$ there is $a_{2k-1}=0<a_{2k}=\frac{1}{(2k{)}^2}$.

So the alternating series test does not apply. However, the series converges by direct comparison:

Exercise

Prove that $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ converges.

Solution

There is

• $|(-1{)}^{k+1}{b}_k|\le \frac{1}{k^2}$ for all $k\in \mathbb{N}$
• $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}<\mathrm{\infty }$

So the series converges by direct comparison, even though its elements do not form a monotone sequence.

Conclusion: Error bounds for the limit

The alternating series test can show converges, but does not give us the limit. For instance, for the alternating harmonic series , there is $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k}=\ln (2)$ . But this limit can not be computed by the alternating series test. However, we can approximate the limit by considering partial sums, and the alternating series test will provide us with a neat upper bound for the error of such an approximation.

We have seen above in this article, that the sequence of partial sums with odd index $(S_{2n-1})$ is monotonically decreasing and converges to the limit $\underset{n\to \mathrm{\infty }}{lim}{S}_{2n-1}=S$ . Further, $S=inf\{S_{2n-1}:n\in \mathbb{N}\}$, where the infimum of a set is the greatest possible lower bound to its elements. Hence, $S_{2n-1}\ge S$ for all $n\in \mathbb{N}$, so we have upper bounds for the limit getting better and better. Conversely, $(S_{2n})$ is monotonically increasing with $\underset{n\to \mathrm{\infty }}{lim}{S}_{2n}=S=sup\{S_{2n}:n\in \mathbb{N}\}$ . So $S_{2n}\le S$ gives a lower bound for all $n\in \mathbb{N}$. That means, we have an estimate $S_{2n}\le S\le S_{2n-1}$ and $S_{2n}\le S\le S_{2n+1}$.

How good is the estimate? We subtract the two inequalities and get

$\begin{array}{cc}{S}_{2n-1}-S& \le S_{2n-1}-S_{2n}\underset{\text{above}}{\stackrel{\text{see}}{=}}{b}_{2n},\text{ and}\\ S-S_{2n}& \le S_{2n+1}-S_{2n}\stackrel{\text{analogously}}{=}{b}_{2n+1}.\end{array}$

So, the series elements serve as a precision indicator for the estimate of the limit by partial sums:

$|S-S_n|=|\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k-\sum _{k=1}^n(-1{)}^{k+1}{b}_k|\le b_{n+1}.$

Theorem (Error estimate for approximating alternating series)

If an alternating series $\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k$ converges by the alternating series test, then the limit can be approximated by the partial sums with maximum error $|\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}{b}_k-\sum _{k=1}^n(-1{)}^{k+1}{b}_k|\le b_{n+1}.$

Example (Error estimate for approximating alternating series)

Let us try to get some numerical values for the limit of the alternating harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k}$:

We take $n=8$ and get the estimate precision $|\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{k}-\sum _{k=1}^8\frac{(-1{)}^{k+1}}{k}|\le b_9=\frac{1}{9}$. Since $S_8\le S$ , we know that the limit must be in the interval $[\frac{533}{840},\frac{533}{840}+\frac{1}{9}]$. We round these fractions to decimal numbers and obtain $S\in [0.634,0.746]$. In fact, the limit is $S=\ln (2)\approx 0.693$. However, this is only a coarse approximation, which already needed summing up 8 terms. There are better ways to compute the logarithm, e.g. Newton's method. The good news is that the above estimate works for any series which converges by the alternating series test.

Generalizing the alternating series test to the Dirichlet test

The Dirichlet test serves for proving convergence of series of the form $\sum _{k=1}^{\mathrm{\infty }}{a}_k{b}_k$ . It extends the alternating series test to cases where there is not $a_k=(-1{)}^{k+1}$ . This is particularly useful, if the sign does not change from element to element (like $+,-,+,-,+,-,+,\mathrm{...}$) but can have streaks without a change in between (like $+,-,-,+,-,+,+,\mathrm{...}$) . The proof is based on Abel's partial summation, which is quite some work to do. We will not state it here.

Theorem (Dirichlet's test)

Let $(a_k)$ and $(b_k)$ be real sequences with

• The partial sum sequence $A_n=\sum _{k=1}^n{a}_k$ being bounded and
• $(b_k)$ being monotonically decreasing
• $\underset{k\to \mathrm{\infty }}{lim}{b}_k=0$.

Then, the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k{b}_k$ converges.

The conditions for $(b_k)$ are exactly the same as for the alternating series test. Actually, with $a_k=(-1{)}^{k+1}$, we just get the alternating series test as a special case:

Exercise

Prove that $a_k=(-1{)}^{k+1}$ fulfils the first condition for Dirichlet's test, i.e.$A_n=\sum _{k=1}^n{a}_k$ is bounded.

Solution

There is

$A_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^n(-1{)}^{k+1}=1-1+1-1\pm \dots +(-1{)}^n=\{\begin{array}{cc}1& \text{ for odd }n,\\ 0& \text{ for even }n.\end{array}$

So $(A_n)$ alternates between 1 and 0 and is obviously bounded.