# Differentiation and Integration of Power Series

Let the power series $$\sum\limits_{n = 0}^\infty {{a_n}{x^n}}$$ have the radius of convergence R > 0. Let

$f\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} = {a_0} + {a_1}x + {a_2}{x^2} + \ldots ,\;\; \left| x \right| \lt R.$

Then for |x| < R the function $$f\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}}$$ is continuous. The power series can be differentiated term-by-term inside the interval of convergence. The derivative of the power series exists and is given by the formula

$f^\prime \left( x \right) = \frac{d}{{dx}}{a_0} + \frac{d}{{dx}}{a_1}x + \frac{d}{{dx}}{a_2}{x^2} + \ldots = {a_1} + 2{a_2}x + 3{a_3}{x^2} + \ldots = \sum\limits_{n = 1}^\infty {n{a_n}{x^{n - 1}}} .$

The power series can be also integrated term-by-term on an interval lying inside the interval of convergence. Hence, if $$- R \lt b \lt x \lt R,$$ the following expression is valid:

$\int\limits_b^x {f\left( t \right)dt} = \int\limits_b^x {{a_0}dt} + \int\limits_b^x {{a_1}tdt} + \int\limits_b^x {{a_2}{t^2}dt} + \ldots + \int\limits_b^x {{a_n}{t^n}dt} + \ldots$

If the series is integrated on the interval $$\left[ {0,x} \right],$$ we can write:

$\int\limits_0^x {f\left( t \right)dt} = \int\limits_0^x {{a_0}dt} + \int\limits_0^x {{a_1}tdt} + \int\limits_0^x {{a_2}{t^2}dt} + \ldots + \int\limits_0^x {{a_n}{t^n}dt} + \ldots = {a_0}x + {a_1}\frac{{{x^2}}}{2} + {a_2}\frac{{{x^3}}}{3} + \ldots = \sum\limits_{n = 0}^\infty {{a_n}\frac{{{x^{n + 1}}}}{{n + 1}}} + C.$

## Solved Problems

### Example 1.

Show that

$\frac{1}{{1 + x}} = 1 - x + {x^2} - {x^3} + {x^4} - \ldots = \sum\limits_{n = 0}^\infty {{a_n}{x^n}}$

for $$\left| {x} \right| \lt 1.$$

Solution.

First we consider the power series:

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

This is a geometric series with ratio $$x.$$ Therefore, it converges for $$\left| x \right| \lt 1.$$ The sum of the series is $${\frac{1}{{1 - x}}}.$$ Substituting $$-x$$ for $$x,$$ we have

$1 - x + {x^2} - {x^3} + \ldots = \frac{1}{{1 - \left( { - x} \right)}} = \frac{1}{{1 + x}}\;\; \text{for}\;\;\left| x \right| \lt 1.$

Thus,

$\frac{1}{{1 + x}} = 1 - x + {x^2} - {x^3} + {x^4} - \ldots = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{x^n}} \;\; \text{for}\;\;\left| x \right| \lt 1.$

### Example 2.

Find a power series for the rational fraction $\frac{1}{{2 - x}}.$

Solution.

We can write this function as

$\frac{1}{{2 - x}} = \frac{{\frac{1}{2}}}{{1 - \frac{x}{2}}}.$

As you can see, this is the sum of the infinite geometric series with the first term $${\frac{1}{2}}$$ and ratio $${\frac{x}{2}}:$$

$\frac{1}{2} + \frac{1}{2}\frac{x}{2} + \frac{1}{2}{\left( {\frac{x}{2}} \right)^2} + \frac{1}{2}{\left( {\frac{x}{2}} \right)^3} + \ldots = \frac{1}{2} + \frac{x}{{{2^2}}} + \frac{{{x^2}}}{{{2^3}}} + \frac{{{x^3}}}{{{2^4}}} + \ldots = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{{2^{n + 1}}}}} .$

The given power series converges for $$\left| x \right| \lt 2.$$

See more problems on Page 2.