Let the power series \(\sum\limits_{n = 0}^\infty {{a_n}{x^n}}\) have the radius of convergence \(R \gt 0.\) Let
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {{a_n}{x^n}} = {a_0} + {a_1}x + {a_2}{x^2} + \ldots ,\;\; \left| x \right| \lt R.\]
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
\[f^\prime \left( x \right) = \frac{d}{{dx}}{a_0} + \frac{d}{{dx}}{a_1}x + \frac{d}{{dx}}{a_2}{x^2} + \ldots = {a_1} + 2{a_2}x + 3{a_3}{x^2} + \ldots = \sum\limits_{n = 1}^\infty {n{a_n}{x^{n - 1}}} .\]
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
Click or tap a problem to see the solution.
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 \(\left| {x} \right| \lt 1.\)
Example 2
Find a power series for the rational fraction \[\frac{1}{{2 - x}}.\]
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 \(\left| x \right| \lt 2.\)
