# Calculus

## Infinite Sequences and Series # Comparison Tests

• The $$N$$th 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 $$0 \lt {a_n} \le {b_n}$$ 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 $$0 \lt \lim\limits_{n \to \infty } {\large\frac{{{a_n}}}{{{b_n}}}\normalsize} \lt \infty ,$$ then $$\sum\limits_{n = 1}^\infty {{a_n}}$$ and $$\sum\limits_{n = 1}^\infty {{b_n}}$$ are both convergent or both divergent;
• If $$\lim\limits_{n \to \infty } {\large\frac{{{a_n}}}{{{b_n}}}\normalsize} = 0,$$ then $$\sum\limits_{n = 1}^\infty {{b_n}}$$ convergent implies that the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is also convergent;
• If $$\lim\limits_{n \to \infty } {\large\frac{{{a_n}}}{{{b_n}}}\normalsize} = \infty,$$ 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 $$\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^p}}}\normalsize}$$ converges for $$p \gt 1$$ and diverges for $$0 \lt p \le 1.$$

• ## Solved Problems

Click a problem to see the solution.

### Example 1

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

### Example 2

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

### Example 3

Determine whether the series $$\sum\limits_{n = 1}^\infty {\large\frac{{{n^2}}}{{{n^3} – 3}}\normalsize}$$ converges or diverges.

### Example 4

Determine whether the series $$\sum\limits_{n = 1}^\infty {\large\frac{{3n – 1}}{{2{n^3} – 4n + 5}}\normalsize}$$ converges or diverges.

### Example 5

Determine whether $$\sum\limits_{n = 1}^\infty {\large\frac{{\sqrt n }}{{2{n^2} + n + 5}}\normalsize}$$ is convergent.

### Example 6

Determine whether the series $$\sum\limits_{n = 1}^\infty {\large\frac{n}{{{n^2} – 2n – 3}}\normalsize}$$ converges or diverges.

### Example 7

Determine whether the series
${\frac{1}{{\sqrt 2 }} + \frac{1}{{2\sqrt 3 }} }+{ \frac{1}{{3\sqrt 4 }} + \ldots }+{ \frac{1}{{n\sqrt {n + 1} }} + \ldots }$
converges or diverges.

### Example 1.

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

Solution.

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

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

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

### Example 2.

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

Solution.

We use the comparison test. Note that $${\large\frac{{{n^2} – 1}}{{{n^4}}}\normalsize}$$ $${\large\frac{{{n^2}}}{{{n^4}}}\normalsize}$$ $$= {\large\frac{1}{{{n^2}}}\normalsize}$$ for all positive integers $$n.$$ As $$\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^2}}}\normalsize}$$ is a $$p$$-series with $$p = 2 \gt 1$$, it converges. Hence, the given series also converges by the comparison test.

Page 1
Problems 1-2
Page 2
Problems 3-7