Select Page

# Calculus

Infinite Sequences and Series

# Power Series

Page 1
Problems 1-2
Page 2
Problems 3-6

### Definition

A series, terms of which are power functions of variable $$x,$$ is called the power series:

${\sum\limits_{n = 1}^\infty {{a_n}{x^n}} } = {{a_0} + {a_1}x }+{ {a_2}{x^2} + \ldots }+{ {a_n}{x^n} + \ldots }$

A series in $$\left( {x – {x_0}} \right)$$ is also often considered. This power series is written as

${\sum\limits_{n = 1}^\infty {{a_n}{{\left( {x – {x_0}} \right)}^n}} } = {{a_0} + {a_1}\left( {x – {x_0}} \right) }+{ {a_2}{\left( {x – {x_0}} \right)^2} + \ldots } + {{a_n}{\left( {x – {x_0}} \right)^n} + \ldots ,}$

where $${x_0}$$ is a real number.

### The Interval and Radius of Convergence

Consider the function $$f\left( x \right) =$$ $$\sum\limits_{n = 1}^\infty {{a_n}{{\left( {x – {x_0}} \right)}^n}}.$$ The domain of this function is the set of those values of $$x$$ for which the series is convergent. The domain of such function is called the interval of convergence.

If the interval is $$\left( {{x_0} – R,{x_0} + R} \right)$$ for some $$R \gt 0,$$ (together with one or both of the endpoints), the $$R$$ is called the radius of convergence. Convergence of the series at the endpoints is determined separately.

Using the root test, the radius of convergence is given by the formula

$R = \lim\limits_{n \to \infty } \frac{1}{{\sqrt[\large n\normalsize]{{{a_n}}}}}$

but a fast way to compute it is based on the ratio test:

$R = \lim\limits_{n \to \infty } \left| {\frac{{{a_n}}}{{{a_{n + 1}}}}} \right|.$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Find the radius of convergence and interval of convergence of the power series $$\sum\limits_{n = 0}^\infty {\large\frac{{{{\left( {x + 3} \right)}^n}}}{{n!}}\normalsize}.$$

### ✓Example 2

Determine the radius of convergence and interval of convergence of the power series $$\sum\limits_{n = 0}^\infty {n{x^n}}.$$

### ✓Example 3

Find the radius of convergence and interval of convergence of the series

${\frac{x}{1} + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} + \ldots }+{ \frac{{{x^n}}}{n} + \ldots }$

### ✓Example 4

For what values of $$x$$ does the series $$\sum\limits_{n = 0}^\infty {\large\frac{{{x^n}}}{{n + 1}}\normalsize}$$ converge?

### ✓Example 5

Find the radius of convergence and interval of convergence of the power series $$\sum\limits_{n = 0}^\infty {\large\frac{{{{\left( { – 1} \right)}^n}{{\left( {x – 2} \right)}^n}}}{{{n^2}}}\normalsize}.$$

### ✓Example 6

Find the radius of convergence and interval of convergence of the power series

${1 + \frac{{2x}}{{\sqrt {5 \cdot 5} }} }+{ \frac{{4{x^2}}}{{\sqrt {9 \cdot {5^2}} }} }+{ \frac{{8{x^3}}}{{\sqrt {13 \cdot {5^3}} }} + \ldots }$

### Example 1.

Find the radius of convergence and interval of convergence of the power series $$\sum\limits_{n = 0}^\infty {\large\frac{{{{\left( {x + 3} \right)}^n}}}{{n!}}\normalsize}.$$

#### Solution.

We make the substitution: $$u = x + 3.$$ The series becomes $$\sum\limits_{n = 0}^\infty {\large\frac{{{u^n}}}{{n!}}\normalsize}.$$ Calculate the radius of convergence:

${R = \lim\limits_{n \to \infty } \left| {\frac{{{a_n}}}{{{a_{n + 1}}}}} \right| } = {\lim\limits_{n \to \infty } \frac{{\frac{1}{{n!}}}}{{\frac{1}{{\left( {n + 1} \right)!}}}} } = {\lim\limits_{n \to \infty } \frac{{\left( {n + 1} \right)!}}{{n!}} } = {\lim\limits_{n \to \infty } \left( {n + 1} \right) }={ \infty .}$

Then the interval of convergence is $$\left( { – \infty ,\infty } \right).$$

### Example 2.

Determine the radius of convergence and interval of convergence of the power series $$\sum\limits_{n = 0}^\infty {n{x^n}}.$$

#### Solution.

${R = \lim\limits_{n \to \infty } \left| {\frac{{{a_n}}}{{{a_{n + 1}}}}} \right| } = {\lim\limits_{n \to \infty } \frac{n}{{n + 1}} } = {\lim\limits_{n \to \infty } \frac{1}{{1 + \frac{1}{n}}} }={ 1.}$

Consider convergence at the endpoints.

If $$x = -1,$$ we have the divergent series $$\sum\limits_{n = 0}^\infty {{{\left( { – 1} \right)}^n}n}.$$

If $$x = 1,$$ the series $$\sum\limits_{n = 0}^\infty n$$ is also divergent.

Therefore, the initial series $$\sum\limits_{n = 0}^\infty {n{x^n}}$$ converges in the open interval $$\left( { -1, 1} \right).$$

Page 1
Problems 1-2
Page 2
Problems 3-6