Deciding convergence by comparison to a series already known to converge or diverge

The comparison test is an easy test for convergence or divergence when we know that, term by term, the terms of a test series are either greater than or less than those of our series of interest.

Here is a statement of the test in two parts, then we'll do some examples to illustrate it:

Example 1

Does the series   $\sum_{n = 1}^\infty \,\frac{1}{n!}$   converge?

Solution: When we look at this series, it's how rapidly the denominator grows that's important. We know that the series

$$\sum_{n = 1}^\infty \,\frac{1}{n^2}$$

the harmonic series, diverges, but that the p-series

$$\sum_{n = 1}^\infty \,\frac{1}{n}$$

converges. Now if each term of the series of interest is greater than $1/n$ for $n \gt 1$ then the series diverges, and if each term is less than $1/n^2$, then the series converges by the comparison test.

We know that for $n\ge 4$, $n!\gt n^2$, so each term of the series with $a_n=1/n!$ is less than the corresponding term of our convergent p-series, therefore the series converges. Here's a graphical look at these three series, $1/n$ in green, $1/n^2$ in magenta and $1/n!$ in blue. Notice that for $n\ge 4$, $1/n!$ is smaller, term-by-term, than $1/n^2$.

Example 2

Does the series   $\sum_{n = 1}^\infty \, \frac{\sqrt{n}}{n^2 + 5}$   converge?

Solution: For a series like this, it's helpful to reduce it by approximation. In the first approximation, the 5 in the denominator doesn't really matter that much for large $n$, so let's drop it to make an approximation of the series:

$$\sum_{n = 1}^\infty \, \frac{\sqrt{n}}{n^2} = \sum_{n = 1}^\infty \frac{n^{1/2}}{n^2} = \sum_{n = 1}^\infty \,\frac{1}{n^{3/2}}$$

Now the series with terms $a_n = 1/n^{3/2}$ is a convergent p-series ($p = 3/2 \gt 1$). That suggests that we can show that each of the terms of our series is smaller than the corresponding term of the p-series with $p = 3/2$. First we set up the trial inequality:

$$\frac{\sqrt{n}}{n^2 + 5} \lt \frac{1}{n^{3/2}}$$

Cross multiplication gives:

$$n^{3/2} n^{1/2} \lt n^2 + 5$$

Using the laws of exponents on the left, we get:

$$n^2 \lt n^2 + 5$$

which reduces to

$$0 \lt 5$$

Because $0 \lt 5$ is always true, our series is, term-by-term, smaller than the convergent p-series with $a_n = 1/n^{3/2}$, so it converges.

Here (right column, top) is a table of the first ten terms of our series and the comparison series:

Example 3

Does the series   $\sum_{n = 1}^\infty \,\frac{1}{n^2} cos \left(\frac{1}{n} \right)$   converge?

Solution: This series contains an oscillating factor, $\text{cos}(1/n)$. Sometimes that factor is positive and sometimes it's negative. Now $1/n$ approaches zero as $n$ gets large, so the cosine part approaches 1. The important thing is that it won't ever exceed 1.

Example 4

Does the series   $\sum_{n = 1}^\infty \,\frac{n + 1}{\sqrt{n}}$   converge?

Solution: It's not too difficult to see that this series has to diverge. If we drop the 1 in the numerator, each term is just $n^{1/2}$, and the sum will grow infinitely large. But let's do it by comparison with the divergent p-series with terms $a_n=1/n^{1/2}$.

First let's break the sum into two parts.

Practice problems

Determine whether these series converge using the comparison test.

1. $$\sum_{n = 2}^\infty \, \frac{n^3}{n^4 - 1}$$

   Solution

   
   

   $$\sum_{n = 2}^\infty \, \frac{n^3}{n^4 - 1}$$

   This series looks like the harmonic series, and the -1 in the denominator makes it look like it's greater, term-by-term, than the terms of that series. Let's compare them:

   $$\frac{n^3}{n^4 - 1} \stackrel{?}{\gt} \frac{1}{n}$$

   $$\longrightarrow \; n^4 \gt n^4 - 1$$

   This is true for all n, so the series diverges; all of its terms are greater than the terms of the divergent harmonic series.




2. $$\sum_{n = 1}^\infty \, \frac{e^{1/n}}{n}$$

   Solution

   
   

   $$\sum_{n = 1}^\infty \, \frac{e^{1/n}}{n}$$

   This series looks like the harmonic series, and the e^1/n term in the numerator makes it look like it's greater, term-by-term, than that series. Let's compare

   $$\frac{e^{1/n}}{n} \stackrel{?}{\gt} \frac{1}{n}$$

   $$ \begin{align} \longrightarrow \; ne^{1/n} &\gt n \\[5pt] e^{1/n} &\gt 1 \\[5pt] e &\gt 1^n \end{align}$$

   This is true for all $n$, so the series diverges; all of its terms are greater than the terms of the divergent harmonic series.




3. $$\sum_{n = 0}^\infty \, \frac{n + 5^n}{n + 6^n}$$

   Solution

   
   

   $$\sum_{n = 0}^\infty \, \frac{n + 5^n}{n + 6^n}$$

   When n is large, the terms of the form "n" divide to 1. What's left is basically $\left( \frac{5}{6} \right)^n,$ which is a convergent geometric series. So let's try to show that, term-by-term, the terms of this series are less than those of that geometric series:

   $$\frac{n + 5^n}{n+ 6^n} \stackrel{?}{\lt} \left( \frac{5}{6}\right)^n$$

   $$ \begin{align} \longrightarrow \; n 6^n + 5^n6^n &\lt n 5^n + 6^{2n} \\[5pt] n(6^n - 5^n) &\lt 6n(6^n - 5^n) \\[5pt] n &\lt 6n \end{align}$$

   This is true for all n > 0, so the series converges