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

\[
{f\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} }
= {{a_0} + {a_1}x }+{ {a_2}{x^2} + \ldots ,\;\;}\kern-0.3pt{\left| x \right| \lt R.}
\]

Then for \(\left| x \right| \lt R\) 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

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

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

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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 \({\large\frac{1}{{1 – x}}\normalsize}.\) Substituting \(-x\) for \(x,\) we have

\[
{1 – x + {x^2} }-{ {x^3} + \ldots }
= {\frac{1}{{1 – \left( { – x} \right)}} }
= {\frac{1}{{1 + x}}\;\;}\kern-0.3pt{\text{for}\;\;\left| x \right| \lt 1.}
\]

Thus,

\[
{\frac{1}{{1 + x}} = 1 – x + {x^2} }-{ {x^3} + {x^4} – \ldots }
= {\sum\limits_{n = 0}^\infty {{{\left( { – 1} \right)}^n}{x^n}} \;\;}\kern-0.3pt
{\text{for}\;\;\left| x \right| \lt 1.}
\]

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 \({\large\frac{1}{2}\normalsize}\) and ratio \({\large\frac{x}{2}\normalsize}:\)

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

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