Direct comparison test – "Math for Non-Geeks"

In this chapter we will study an important convergence criterion called the direct comparison test. This criterion allows us to deduce the convergence behavior of a series by comparing it to that of another series. That way we can answer questions regarding the convergence of a series by considering a simpler series. Using this criterion we can estimate upper or lower bounds of the series, until we hopefully find a proof for the convergence behavior.

The direct comparison test is also used to prove other criteria such as the quotient test and the root test, which are both useful tools to have when solving problems about the convergence of series.

Direct comparison test: majorant

There are two versions of the direct comparison test. The first one is a test for convergence and involves finding a majorant of the series:

Theorem (Direct comparison: majorant)

Let $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ be a series. If there exists a convergent series $\sum _{k=1}^{\mathrm{\infty }}{c}_k$ with $|a_k|\le c_k$ for all $k\in \mathbb{N}$ then $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges absolutely.

Note that from the inequality $|a_k|\le c_k$ it automatically follows that $c_k\ge 0$, because $c_k$ is greater or equal to the non-negative number $|a_k|$.

Proof (Direct comparison: majorant)

If $\sum _{k=1}^{\mathrm{\infty }}{c}_k$ converges, then the sequences of partial sums is bounded above . Because $|a_k|\le c_k$ for all $k\in \mathbb{N}$ also the corresponding sequence of partial sums of $\sum _{k=1}^{\mathrm{\infty }}|a_k|$ has an upper bound:

$\sum _{k=1}^n|a_k|\le \sum _{k=1}^n{c}_k\le \sum _{k=1}^{\mathrm{\infty }}{c}_k<\mathrm{\infty }$

The sequence of partial sums $\sum _{k=1}^{\mathrm{\infty }}|a_k|$ is monotonically increasing since $|a_k|\ge 0$. We have that the series is monotonic and bounded. Therefore $\sum _{k=1}^{\mathrm{\infty }}|a_k|$ is convergent.

Comprehension question: Would it suffice for the majorant if there exists $N\in \mathbb{N}$ so that $|a_k|\le c_k$ for all $k\ge N$?

Yes. If the sequence of partial sums $\sum _{k=1}^{\mathrm{\infty }}{c}_k$ is bounded above, then this is true also for $\sum _{k=N}^{\mathrm{\infty }}{c}_k$. Because of

$\sum _{k=N}^n|a_k|\le \sum _{k=N}^n{c}_k\le \sum _{k=N}^{\mathrm{\infty }}{c}_k<\mathrm{\infty }$

also the sequence of partial sums $\sum _{k=N}^{\mathrm{\infty }}|a_k|$ is bounded above. Since the finitely many summands $a_1,a_2,\dots ,a_{N-1}$ don't change the boundedness, the partial sums of $\sum _{k=1}^{\mathrm{\infty }}|a_k|$ also have an upper bound. This means that $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges absolutely.

Hint

As we already mentioned in the introduction, we would like to find a preferably simple convergent series to use as our majorant. Often, we can use the series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}$ as our majorant. We could also use a more general harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$ for $\alpha >1$. Another good option is the convergent geometric series $\sum _{k=0}^{\mathrm{\infty }}{q}^k$ for $|q|<1$, for example $\sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^k$.

Direct comparison test: minorant

The second comparison test is similar, but we use it to determine the divergence of a series, and it involves finding a minorant of that series.

Theorem (Direct comparison test: minorant)

Let $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ be a series with $a_k\ge 0$ for all $k\in \mathbb{N}$. If there exists a divergent series $\sum _{k=1}^{\mathrm{\infty }}{c}_k$ with $a_k\ge c_k\ge 0$ for all $k\in \mathbb{N}$, then also the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ is divergent.

Proof (Direct comparison test: minorant)

Since $c_k\ge 0$ the sequence of partial sums of $\sum _{k=1}^{\mathrm{\infty }}{c}_k$ is monotonic. According to our premise this series is divergent, so it must be unbounded. Because $c_k\le a_k$ for all $k\in \mathbb{N}$ we have that $\sum _{k=1}^n{c}_k\le \sum _{k=1}^n{a}_k$ for all $n\in \mathbb{N}$ and therefore also the sequence of partial sums of $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ is unbounded. This implies that $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ diverges (every unbounded sequence diverges.)

Hint

Analogue to the majorant condition it suffices that $a_k\ge c_k\ge 0$ be true for all $k\ge N$, for a fixed $N\in \mathbb{N}$.

Hint

When using the minorant criterion in practice, often a good choice for the minorant is the divergent harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$. But also the series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{\alpha }}$ for $\alpha <1$ (for example $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\sqrt{k}}$) make for good minorants. Also the geometric series $\sum _{k=1}^{\mathrm{\infty }}{q}^k$ with $q\ge 1$ is suited as a minorant.

Warning

When applying the direct comparison test using a minorant it is necessary that $a_k\ge 0$! If we only have that $|a_k|\ge c_k$ for $a_k\in ℝ$ as in the majorant case, then we cannot follow the divergence of $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ from the divergence of $\sum _{k=1}^{\mathrm{\infty }}{c}_k$. We can only say that $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ doesn't converge absolutely. To see this consider for example the series $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k}{k}$ and $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ with $a_k=\frac{(-1{)}^k}{k}$ and $c_k=\frac{1}{k}$. We have $\frac{1}{k}\le \left|\frac{(-1{)}^k}{k}\right|=\frac{1}{k}$, and the harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ diverges. But we can show that $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^k}{k}$ converges according to the alternating series test.

Examples and exercises

Direct comparison with majorant

Exercise (Direct comparison with majorant)

Examine whether the series $\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{k+1}{k^2+3k}\right)}^2$ converges.

How to get to the proof? (Direct comparison with majorant)

If one has little experience proving convergence of series, it is difficult to tell whether the series is convergent or divergent. Start by examining the fraction $\frac{k+1}{k^2+3k}$. Maybe you see that the numerator is a polynomial of degree two and the denominator is a polynomial of degree three. This means that the whole fraction goes to zero with a rate of convergence $\frac{1}{k}$. In our series this fraction is squared. This means that ${\left(\frac{k+1}{k^2+3k}\right)}^2$ will go to zero with a convergences rate of $\frac{1}{k^2}$. Since $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}$ is convergent, also the series $\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{k+1}{k^2+3k}\right)}^2$ should converge.

To prove this formally, we can use the direct comparison test and compare $\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{k+1}{k^2+3k}\right)}^2$ to the majorant $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}$. But we first need to show that this is indeed a majorant of our series. Here we need to make some clever estimates:

$\begin{array}{cc}\frac{k+1}{k^2+3k}& \le \frac{k+1}{k^2}\le \frac{k+k}{k^2}\\ & =\frac{2k}{k^2}=2\cdot \frac{1}{k}\end{array}$

To estimate an upper bound we followed a schema: We removed summands that make the term smaller. For the remaining summands, that we could not simply remove, we found an estimate so that we could merge them. The end result:

${\left(\frac{k+1}{k^2+3k}\right)}^2\le {(2\cdot \frac{1}{k})}^2=4\cdot \frac{1}{k^2}$

Thus $\sum _{k=1}^{\mathrm{\infty }}4\cdot \frac{1}{k^2}$ is indeed a majorant, and since the series is convergent, by direct comparison we find that also $\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{k+1}{k^2+3k}\right)}^2$ must be convergent.

Solution (Direct comparison with majorant)

We have that:

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

and therefore

$\sum _{k=1}^{\mathrm{\infty }}4\cdot \frac{1}{k^2}=4\cdot \sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}<\mathrm{\infty }$

is a convergent majorant of the series $\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{k+1}{k^2+3k}\right)}^2$. By direct comparison the series is convergent.

Direct comparison with minorant

An example for the direct comparison test (Youtube video by the Youtube channel Maths CA).

Exercise (Direct comparison with minorant)

Examine whether the series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\sqrt{k(k+2)}}$ converges or diverges.

How to get to the proof? (Direct comparison with minorant)

Here it is interesting to note the rate at which the summands converge to zero. The product $k(k+2)$ converges like $k^2$ towards infinity. Thus $\frac{1}{k(k+2)}$ converges to 0 like $\frac{1}{k^2}$. Since we take the square root we can see that the rate of convergence of $\frac{1}{\sqrt{k(k+2)}}$ is like $\frac{1}{k}$. But the harmonic series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ is divergent. This tells us that $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\sqrt{k(k+2)}}$ should also diverge.

Now that we have conjectured the divergence of the series, we need to prove it. We do that by direct comparison with a minorant. First, we need to estimate lower bounds:

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

As in the previous exercise when we needed to estimate an upper bound, we can follow a schema: Summands that make the term bigger, can be removed. For the remaining summands, we try to find an estimate so that we can merge them. So now we have a divergent minorant: $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\sqrt{3}\cdot k}$. By direct comparison, it follows that $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\sqrt{k(k+2)}}$ is divergent.

Proof (Direct comparison with minorant)

We have

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

The series $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\sqrt{3}}\cdot \frac{1}{k}$ is divergent. Thus, also $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\sqrt{k(k+2)}}$ must diverge by direct comparison.

Corollary: Limit comparison test

For series with positive summands there is another test called the limit comparison test, which we can derive from the direct comparison test:

Theorem (Limit comparison test)

Let $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ and $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ be series with positive summands, and $\underset{k\to \mathrm{\infty }}{lim}\frac{a_k}{b_k}=c$ with $c>0$, then the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges if and only if $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ converges.

Proof (Limit comparison test)

Because $\underset{k\to \mathrm{\infty }}{lim}\frac{a_k}{b_k}=c$ there exists $N\in \mathbb{N}$, so that $|\frac{a_k}{b_k}-c|<\frac{c}{2}$ for all $k\ge N$. This follows from the epsilon definition of convergence with $ϵ=\frac{c}{2}$. For all $k\ge N$ we have:

$\begin{array}{cc}& |\frac{a_k}{b_k}-c|<\frac{c}{2}\\ ⟺& c-\frac{c}{2}\le \frac{a_k}{b_k}\le c+\frac{c}{2}\\ ⟺& \frac{1}{2}c\le \frac{a_k}{b_k}\le \frac{3}{2}c\\ ⟺& \frac{1}{2}c\cdot b_k\le a_k\le \frac{3}{2}c\cdot b_k\end{array}$

If $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges, then also $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{2}c\cdot b_k$ converges by direct comparison because of the inequality $\frac{1}{2}c\cdot b_k\le a_k$ for all $k\ge N$. We also see that $\sum _{k=1}^{\mathrm{\infty }}{b}_k=\sum _{k=1}^{\mathrm{\infty }}\frac{2}{c}\cdot (\frac{1}{2}c\cdot b_k)$. But then also $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ converges, because multiplying by a constant factor doesn't affect convergence behaviour.

On the other side assume that $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ is convergent, then also $\sum _{k=1}^{\mathrm{\infty }}\frac{3}{2}c\cdot b_k$ is convergent (again multiplication by a constant factor). Because of the inequality $a_k\le \frac{3}{2}c\cdot b_k$ for all $k\ge N$ the series $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ is convergent by direct comparison.

Comprehension question: What can we deduce from $\underset{k\to \mathrm{\infty }}{lim}\frac{a_k}{b_k}=0$ under the same conditions?

From the convergence of $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ we can deduce the convergence of $\sum _{k=1}^{\mathrm{\infty }}{a}_k$. If $\underset{k\to \mathrm{\infty }}{lim}\frac{a_k}{b_k}=0$, there exists $N\in \mathbb{N}$ with $\frac{a_k}{b_k}\le 1⟺a_k\le b_k$ for all $k\ge N$. Thus, if $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ is convergent it follows from direct comparison that $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ is convergent.

The opposite is not true: If $\sum _{k=1}^{\mathrm{\infty }}{a}_k$ converges we cannot conclude that $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ is also convergent. As a counterexample consider $\sum _{k=1}^{\mathrm{\infty }}\underset{=a_k}{\underbrace{\frac{1}{k^2}}}$ and $\sum _{k=1}^{\mathrm{\infty }}\underset{=b_k}{\underbrace{\frac{1}{k}}}$ . Here we have

$\underset{k\to \mathrm{\infty }}{lim}\frac{a_k}{b_k}=\underset{k\to \mathrm{\infty }}{lim}\frac{\frac{1}{k^2}}{\frac{1}{k}}=\underset{k\to \mathrm{\infty }}{lim}\frac{1}{k}=0$

But $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}$ is convergent and $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k}$ is divergent .

Example (Prove convergence using limit comparison test)

Consider the series $\sum _{k=1}^{\mathrm{\infty }}\frac{2k^2}{6k^4-3k}$. We have

$a_k=\frac{2k^2}{6k^4-3k}=\frac{k^2}{k^4}\cdot \frac{2}{6-\frac{3}{k^3}}=\frac{1}{k^2}\cdot \frac{2}{6-\frac{3}{k^3}}$

With $b_k=\frac{1}{k^2}$ it follows that

$\underset{k\to \mathrm{\infty }}{lim}\frac{a_k}{b_k}=\underset{k\to \mathrm{\infty }}{lim}\frac{2}{6-\frac{3}{k^3}}=\frac{2}{6}=\frac{1}{3}$

Since $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}$ is convergent, is follows from the limit comparison test that $\sum _{k=1}^{\mathrm{\infty }}\frac{2k^2}{6k^4-3k}$ is also convergent .

Hint

As this example shows, the limit comparison test is a convenient way to prove the convergence of a series. Often, computing the limit $\underset{k\to \mathrm{\infty }}{lim}\frac{a_k}{b_k}$ is easier than finding a suiting majorant. In many lectures however the limit comparison test is not discussed, so you are probably not allowed to use it either. Even so you can still use the argumentation of the proof to find the necessary majorant:

For $a_k=\frac{2k^2}{6k^4-3k}$ and $b_k=\frac{1}{k^2}$ we find that $\underset{k\to \mathrm{\infty }}{lim}\frac{a_k}{b_k}=\frac{1}{3}$. There must exist $n_0\in \mathbb{N}$ with $\frac{a_k}{b_k}<2\cdot \frac{1}{3}=\frac{2}{3}$ for all $k\ge n_0$. This is equivalent to $a_k\le \frac{2}{3}{b}_k$ for all $k\ge n_0$. This way we have found a suitable majorant. Because we know that $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}$ converges, also $\sum _{k=1}^{\mathrm{\infty }}\frac{2}{3}\frac{1}{k^2}$ must converge. Now we use the direct comparison test to show that $\sum _{k=1}^{\mathrm{\infty }}\frac{2k^2}{6k^4-3k}$ is convergent.

Exercise (Limit comparison test)

Examine for which $s\in \mathbb{Q}$ the series $\sum _{n=1}^{\mathrm{\infty }}(\sqrt{1+\frac{1}{n^s}}-1)$ converges.

Hint: Consider the general harmonic series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^s}$.

Solution (Limit comparison test)

Case 1: $s\le 0$

Here $a_n=\sqrt{1+\frac{1}{n^s}}-1$ is not a null sequence:

$a_n=\sqrt{1+\frac{1}{n^s}}-1=\{\begin{array}{cc}\sqrt{1+\frac{1}{n^0}}-1=\sqrt{2}-1& \text{for }s=0,\\ \sqrt{1+\frac{1}{n^s}}-1=\sqrt{1+n^{-s}}-1\to \mathrm{\infty }& \text{for }s<0⟺-s>0\end{array}$

Case 2: $s>0$

Let $a_n=\sqrt{1+\frac{1}{n^s}}-1$ and $b_n=\frac{1}{n^s}$. We have

$\begin{array}{cc}\frac{a_n}{b_n}& =\frac{\sqrt{1+\frac{1}{n^s}}-1}{\frac{1}{n^s}}\\ & =\frac{(\sqrt{1+\frac{1}{n^s}}-1)(\sqrt{1+\frac{1}{n^s}}+1)}{\frac{1}{n^s}(\sqrt{1+\frac{1}{n^s}}+1)}\\ & =\frac{1+\frac{1}{n^s}-1^2}{\frac{1}{n^s}(\sqrt{1+\frac{1}{n^s}}+1)}\\ & =\frac{\frac{1}{n^s}}{\frac{1}{n^s}(\sqrt{1+\frac{1}{n^s}}+1)}\\ & =\frac{1}{\sqrt{1+\frac{1}{n^s}}+1}\\ \end{array}$

Because of $\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n^s}=0$ and from the limit theorems it follows $\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{1+\frac{1}{n^s}}+1}=\frac{1}{\sqrt{1}+1}=\frac{1}{2}$. According to the limit comparison test, the series $\sum _{n=1}^{\mathrm{\infty }}(\sqrt{1+\frac{1}{n^s}}-1)$ converges if and only if $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^s}$ converges. Thus, we know it converges for $s>1$ and diverges for $0<s\le 1$.

Combining both cases we find that our series is convergent for $s>1$ and divergent for $s\le 1$.