# Applications of the comparison tests

## 1. Introduction

This web page contains some examples of the use of the comparison test. The comparison test is perhaps the most basic and the most useful test of all and will be used many times later on, including in the proofs of some important theorems and other tests for convergence such as the ratio test.

I will illustrate both versions of the comparison test here. In the forms I have presented them, the basic form of the comparison test is more general and applies to a greater number of series, and in all cases where the limit comparison test is used the other form may be used instead. The limit comparison test may be conceptually a little harder to state and use, and the final calculation does involves a limit, but generally speaking the algebraic manipulations required tend to be easier. You can, if you prefer, just learn the first version of the test and use that throughout.

## 2. Examples using the basic form of the test

Example.

The series $∑ n = 1 ∞ 1 ( n + 1 2 ) 2$ converges.

Proof.

Note that, for all $n ∈ ℕ$, $0 ⩽ 1 ( n + 1 2 ) 2 ⩽ 1 n 2$, and the series $∑ n = 1 ∞ 1 n 2$ is an example of a standard convergent series.

Example.

The series $∑ n = 1 ∞ 1 3 n - 1$ diverges.

Proof.

For all $n ∈ ℕ$, we have $1 3 n - 1 > 1 3 · 1 n$, and the constant $1 3$ is positive and the series $∑ n = 1 ∞ 1 n$ is an example of a standard divergent series.

Example.

The series $∑ n = 1 ∞ n - p$ diverges for all rational $0 ⩽ p < 1$.

Proof.

Again, as $p < 1$ we have $n - p ⩾ 1 n$, so the divergence of the series follows from the comparison test and the divergence of $∑ n = 1 ∞ n -1$.

Example.

Let $x$ be a positive real number. Then the series $∑ n = 1 ∞ x n n$ diverges for $x ⩾ 1$, and converges for $0 < x < 1$.

Proof.

In the case $x = 1$, the series is the harmonic series so diverges, by previous results. If $x > 1$ we have $x n n > 1 n > 0$ for each $n$ and so $∑ n = 1 ∞ x n n$ also diverges, by the comparison test.

If $0 < x < 1$ then $0 < x n n < x n$ for all $n ⩾ 2$ and so by comparison with the convergent geometric series $∑ n = 1 ∞ x n$ the series $∑ n = 1 ∞ x n n$ converges.

## 3. Examples using the limit form of the test

Example.

The series $∑ n = 1 ∞ n 2 + 3 n + 1 n 4 + 3 n 2 + 4 n + 1$ converges.

Proof.

Let $u n = n 2 + 3 n + 1 n 4 + 3 n 2 + 4 n + 1$ and $v n = 1 n 2$. (We chose $v n$ because, by counting indices, we guess that $u n$ behaves like $1 n 2$ as $n → ∞$.) Now compute the limit:

$u n v n = n 2 ( n 2 + 3 n + 1 ) n 4 + 3 n 2 + 4 n + 1 = 1 + 3 n -1 + n -2 1 + 3 n -2 + 4 n -3 + n -4 → 1$

by continuity of $+,·$ and of division at $( 1 , 1 )$, and by standard null sequences $n -1$, $n -2$, etc.

Now as the limit is $1 > 0$ and both series are of positive terms only, by the limit comparison test $∑ n = 1 ∞ u n$ and $∑ n = 1 ∞ v n$ either both diverge or both converge. But $∑ n = 1 ∞ v n = ∑ n = 1 ∞ n -2$ is a standard convergent series and therefore our series also converges.

Example.

The series $∑ n = 1 ∞ n 2 + 3 n + 1 8 n 3 + 3 n 2 + 4 n + 1$ diverges.

Proof.

We perform limit comparision of our series with the harmonic series $∑ n = 1 ∞ 1 n$. We have,

$n ( n 2 + 3 n + 1 ) 8 n 3 + 3 n 2 + 4 n + 1 = 1 + 3 n -1 + n -2 8 + 3 n -1 + 4 n -2 + n -3 → 1 8 ≠ 0$

by continuity and standard null sequences $n -1$, $n -2$, etc. Therefore by the limit comparison test, our sequence behaves as the harmonic series does. But the harmonic series is a standard divergent series, so our series also diverges.