# Power Series

## Definition

A series, terms of which are power functions of variable \(x,\) is called the power series:

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

where \({x_0}\) is a real number.

## The Interval and Radius of Convergence

Consider the function

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

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

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

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

### Example 2

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

### Example 1.

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

Solution.

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

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.

Calculate the radius of convergence:

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 \(\left( { -1, 1} \right).\)