Introduction

Welcome to this guide on Comparison Test, this article will guide you through the following:

Direct Comparison Test (DCT)

Let $a_n$ and $b_n$ be two infinite series, if $0<a_n\le b_n$ for all $n>0$:

Sidenote about DCT

We said before that to apply the DCT, $0<a_n\le b_n$ must hold true for all $n$, but this is a little bit too strict. In fact, the requirement of $a_n\le b_n$ only needs to met eventually for a sufficiently large $n$. Here is what is actually means:

Consider two infinite series $a_n$ and $b_n$. Let us also consider an arbitrary number $N$ such that $a_n\le b_n$ for $n>N$. Let us break down the series in this manner:

$\sum _{n=1}^{\infty }{a}_n=\sum _{n=1}^N{a}_n+\sum _{n=N+1}^{\infty }{a}_n\text{ and }\sum _{n=1}^{\infty }{b}_n=\sum _{n=1}^N{b}_n+\sum _{n=N+1}^{\infty }{b}_n$

Since $N$ is a finite number, $\sum _{n=1}^N{a}_n$ and $\sum _{n=1}^N{b}_n$ are finite amounts. We can apply the DCT on the tails of the series $\left(\sum _{n=N+1}^{\infty }{a}_n\text{ and }\sum _{n=N+1}^{\infty }{b}_n\right)$ and use that to establish the status of the original series.

In general, if the tail of a series converges/diverges, then the entire series must also converge/diverge as well. Thus, if the tail of $b_n$ converges, then the tail of $a_n$ converges, and thus the entirety of $a_n$ converges. Similarly, if the tail of $a_n$ diverges, then the tail of $b_n$ diverges, and thus the entirety of $b_n$ also diverges.

Example 1:

Determine the convergence of this series

$\sum _{n=1}^{\infty }\frac{1}{2+3^n}$

Note that this series looks similar to this series, which is a geometric series that converges:

$\sum _{n=1}^{\infty }\frac{1}{3^n}$

It can be proven that this holds true when $n>1$ (this expression holds true if $n>0$)

$\frac{1}{3^n}>\frac{1}{2+3^n}$

Thus, $a_n=\sum _{n=1}^{\infty }\frac{1}{2+3^n}$ and $b_n=\sum _{n=1}^{\infty }\frac{1}{3^n}$. Since $b_n$ converges, this means that the original series converges.

Example 2:

Determine the convergence of this series:

$\sum _{n=1}^{\infty }\frac{1}{5+\sqrt{n}}$

This series look similar to this series:

$\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}$

Notice that  $\frac{1}{5+\sqrt{n}}<\frac{1}{\sqrt{n}}$ when $n>1$. Thus, $a_n=\sum _{n=1}^{\infty }\frac{1}{5+\sqrt{n}}$ and $b_n=\sum _{n=1}^{\infty }\frac{1}{\sqrt{n}}$. However, $b_n$ diverges as it is a p-series with $p=\frac{1}{2}<1$. If we try to apply DCT here, we notice that we cannot draw any conclusion, because if $b_n$ diverges, then $a_n$'s status is unknown. Thus, we must switch to a new series to solve this problem:

$\sum _{n=1}^{\infty }\frac{1}{n}$

Notice that  $\frac{1}{5+\sqrt{n}}>\frac{1}{n}$ when $n>8$, which means we can now apply the DCT ($N=8$ here). Thus, $a_n=\sum _{n=9}^{\infty }\frac{1}{n}$ and $b_n=\sum _{n=9}^{\infty }\frac{1}{5+\sqrt{n}}$. We know that as the tail of $a_n$ diverges, then the tail and entirety of $b_n$ MUST diverge.

Limit Comparison Test (LCT)

If $a_n>0$ and $b_n>0$, if $\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=L$, there are 3 cases to consider:

If $L=0$ and $b_n$ diverge, do not assume $a_n$ diverges as well. The same also applies for when $L=\infty$.

Example 1:

Determine the convergence of this series:

$\sum _{n=1}^{\infty }\frac{\sqrt{n}}{n^2+1}$

Let's examine this series:

$\sum _{n=1}^{\infty }\frac{\sqrt{n}}{n^2}=\sum _{n=1}^{\infty }\frac{1}{n^{\frac{3}{2}}}$

Which is a convergent p-series. Let us set up $a_n=\sum _{n=1}^{\infty }\frac{\sqrt{n}}{n^2+1}$ and $b_n=\sum _{n=1}^{\infty }\frac{\sqrt{n}}{n^2}$, and now apply the LCT and use L'Hôpital's rule to simplify the limit:

$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}$

$\underset{n\to \infty }{\lim }{a}_n\ast \frac{1}{b_n}$

$\underset{n\to \infty }{\lim }\frac{\sqrt{n}}{n^2+1}\frac{n^2}{\sqrt{n}}$

$=\underset{n\to \infty }{\lim }\frac{n^2}{n^2+1}$

$=1$

Which means the original series converges.

Example 2:

Determine the convergence of this series:

$\sum _{n=1}^{\infty }\frac{n{2}^n}{4n^3+1}$

We construct this series to be the compared series ($b_n$).

$\sum _{n=1}^{\infty }\frac{2^n}{n^2}$

To see if this series converges or not, we use the $n$-th term test (we applied L'Hôpital's Rule to this limit):

$\underset{n\to \infty }{\lim }\frac{2^n}{n^2}=\underset{n\to \infty }{\lim }\frac{\ln 2\cdot 2^n}{2n}$

$=\underset{n\to \infty }{\lim }\frac{\ln 2\cdot \ln 2\cdot 2^n}{2}>0$

This means that the series $b_n$ diverges, so then we apply the LCT:

$\underset{n\to \infty }{\lim }\frac{n\cdot 2^n}{4n^3+1}\frac{n^2}{2^n}$

$=\underset{n\to \infty }{\lim }\frac{n^3}{4n^3+1}$

$=\frac{1}{4}$

Which is a finite positive number, meaning the original series $a_n$ diverges.

Tricks and Tips at selecting the compared series

You should see that in order to apply the comparison test, choosing the right series to compare is the key, here are some tricks for choosing the compared series:

However those are not firm guidelines you should follow, be creative when you do problems!

Practice

Determine if the following series converge using the comparison test: