# Comparison Tests

The Nth term test, generally speaking, does not guarantee convergence of a series. Convergence or divergence of a series is proved using sufficient conditions. The comparison tests we consider below are just the sufficient conditions of convergence or divergence of series.

## The Comparison Tests

Let $$\sum\limits_{n = 1}^\infty {{a_n}}$$ and $$\sum\limits_{n = 1}^\infty {{b_n}}$$ be series such that 0 < anbn for all n. Then the following comparison tests hold:

• If $$\sum\limits_{n = 1}^\infty {{b_n}}$$ is convergent, then $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is also convergent;
• If $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is divergent, then $$\sum\limits_{n = 1}^\infty {{b_n}}$$ is also divergent.

## The Limit Comparison Tests

Let $$\sum\limits_{n = 1}^\infty {{a_n}}$$ and $$\sum\limits_{n = 1}^\infty {{b_n}}$$ be series such that $${a_n}$$ and $${b_n}$$ are positive for all $$n.$$ Then the following limit comparison tests are valid:

• If $$0 \lt \lim\limits_{n \to \infty } {\frac{{{a_n}}}{{{b_n}}}} \lt \infty ,$$ then $$\sum\limits_{n = 1}^\infty {{a_n}}$$ and $$\sum\limits_{n = 1}^\infty {{b_n}}$$ are both convergent or both divergent;
• If $$\lim\limits_{n \to \infty } {\frac{{{a_n}}}{{{b_n}}}} = 0,$$ then $$\sum\limits_{n = 1}^\infty {{b_n}}$$ convergent implies that the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is also convergent;
• If $$\lim\limits_{n \to \infty } {\frac{{{a_n}}}{{{b_n}}}} = \infty,$$ then $$\sum\limits_{n = 1}^\infty {{b_n}}$$ divergent implies that the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is also divergent.

The so-called $$p$$-series $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^p}}}}$$ converges for $$p \gt 1$$ and diverges for $$0 \lt p \le 1.$$

## Solved Problems

### Example 1.

Determine whether $\sum\limits_{n = 1}^\infty {\frac{{{e^{\frac{1}{n}}}}}{{{n^2}}}}$ converges or diverges.

Solution.

We easily can see that $${e^{\frac{1}{n}}} \le e$$ for $$n \gt 1.$$ Then, by the comparison test,

$\sum\limits_{n = 1}^\infty {\frac{{{e^{\frac{1}{n}}}}}{{{n^2}}}} \le \sum\limits_{n = 1}^\infty {\frac{e}{{{n^2}}}} = e\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} .$

Since the series $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}$$ is convergent as a $$p$$-series with the power $$p = 2,$$ the original series also converges.

### Example 2.

Determine whether the series $\sum\limits_{n = 1}^\infty {\frac{{{n^2} - 1}}{{{n^4}}}}$ converges or diverges.

Solution.

We use the comparison test. Note that

$\frac{{{n^2} - 1}}{{{n^4}}} \lt \frac{{{n^2}}}{{{n^4}}} = \frac{1}{{{n^2}}}$

for all positive integers $$n.$$ As $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}}$$ is a $$p$$-series with $$p = 2 \gt 1$$, it converges. Hence, the given series also converges by the comparison test.

See more problems on Page 2.