# Infinite Series

## Definitions

Let $$\left\{ {{a_n}} \right\}$$ be a sequence. Then the infinite sum

$\sum\limits_{n = 1}^\infty {{a_n}} = {a_1} + {a_2} + \ldots + {a_n} + \ldots$

is called an infinite series, or, simply, series. The partial sums of the series are given by

$\sum\limits_{n = 1}^n {{a_n}} = {a_1} + {a_2} + \ldots + {a_n},$

where $${S_n}$$ is called the $$n$$th partial sum of the series. If the partial sums $$\left\{ {{S_n}} \right\}$$ converge to $$L$$ as $$n \to \infty,$$ then we say that the infinite series converges to $$L:$$

$\sum\limits_{n = 1}^\infty {{a_n}} = L,\;\; \text{if}\;\;\lim\limits_{n \to \infty } {S_n} = L.$

Otherwise we say that the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ diverges.

## $$N$$th term test

If the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is convergent, then $$\lim\limits_{n \to \infty } {a_n} = 0.$$

### Important!

The converse of this theorem is false. The convergence of $${{a_n}}$$ to zero does not imply that the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ converges. For example, the harmonic series $$\sum\limits_{n = 1}^\infty {\frac{1}{n}}$$ diverges (see Example $$3$$), although $$\lim\limits_{n \to \infty } {a_n} = 0.$$

Equivalently, if $$\lim\limits_{n \to \infty } {a_n} \ne 0$$ or this limit does not exist, then the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is divergent.

## Properties of Convergent Series

Let $$\sum\limits_{n = 1}^\infty {{a_n}} = A$$ and $$\sum\limits_{n = 1}^\infty {{b_n}} = B$$ be convergent series and let $$c$$ be a real number. Then

• $$\sum\limits_{n = 1}^\infty {\left( {{a_n} + {b_n}} \right)} = A + B$$
• $$\sum\limits_{n = 1}^\infty {c{a_n}} = cA$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Determine whether $\sum\limits_{n = 1}^\infty {\sqrt[n]{3}}$ converges or diverges.

### Example 2

Investigate convergence of the series $\sum\limits_{n = 1}^\infty {\frac{{{e^n}}}{{{n^2}}}}.$

### Example 1.

Determine whether $\sum\limits_{n = 1}^\infty {\sqrt[n]{3}}$ converges or diverges.

Solution.

Since $$\lim\limits_{n \to \infty } \sqrt[n]{3} = \lim\limits_{n \to \infty } {3^{\frac{1}{n}}} = 1,$$ then the series $$\sum\limits_{n = 1}^\infty {\sqrt[n]{3}}$$ diverges by the $$n$$th term test.

### Example 2.

Investigate convergence of the series $\sum\limits_{n = 1}^\infty {\frac{{{e^n}}}{{{n^2}}}}.$

Solution.

Calculate the limit $$\lim\limits_{n \to \infty } {\frac{{{e^n}}}{{{n^2}}}}.$$ Using L'Hopital's rule, we find

$\lim\limits_{x \to \infty } \frac{{{e^x}}}{{{x^2}}} = \lim\limits_{x \to \infty } \frac{{{e^x}}}{{2x}} = \lim\limits_{x \to \infty } \frac{{{e^x}}}{2} = \infty .$

Hence, the original series diverges by the $$n$$th term test.