
# 11.9: Power Series

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Recall that we were able to analyze all geometric series "simultaneously'' to discover that $\sum_{n=0}^\infty kx^n = {k\over 1-x},$ if $$|x| < 1$$, and that the series diverges when $$|x|\ge 1$$. At the time, we thought of $$x$$ as an unspecified constant, but we could just as well think of it as a variable, in which case the series $\sum_{n=0}^\infty kx^n$ is a function, namely, the function $$k/(1-x)$$, as long as $$|x| < 1$$. While $$k/(1-x)$$ is a reasonably easy function to deal with, the more complicated $$\sum kx^n$$ does have its attractions: it appears to be an infinite version of one of the simplest function types---a polynomial. This leads naturally to the questions: Do other functions have representations as series? Is there an advantage to viewing them in this way?

The geometric series has a special feature that makes it unlike a typical polynomial---the coefficients of the powers of $$x$$ are the same, namely $$k$$. We will need to allow more general coefficients if we are to get anything other than the geometric series.

Definition 11.8.1

A power series has the form $$\sum_{n=0}^\infty a_nx^n,$$ with the understanding that $$a_n$$ may depend on $$n$$ but not on $$x$$.

Example 11.8.2

$$\sum_{n=1}^{\infty} x^n \over n$$ is a power series. We can investigate convergence using the ratio test:

$\lim_{n\to\infty} {|x|^{n+1}\over n+1}{n\over |x|^n} =\lim_{n\to\infty} |x|{n\over n+1} =|x|.$

Thus when $$|x| < 1$$ the series converges and when $$|x|>1$$ it diverges, leaving only two values in doubt. When $$x=1$$ the series is the harmonic series and diverges; when $$x=-1$$ it is the alternating harmonic series (actually the negative of the usual alternating harmonic series) and converges. Thus, we may think of

$\sum_{n=1}^\infty {x^n\over n}$

as a function from the interval $$[-1,1])$$ to the real numbers.

A bit of thought reveals that the ratio test applied to a power series will always have the same nice form. In general, we will compute

$\lim_{n\to\infty} {|a_{n+1}||x|^{n+1}\over |a_n||x|^n} =\lim_{n\to\infty} |x|{|a_{n+1}|\over |a_n|} = |x|\lim_{n\to\infty} {|a_{n+1}|\over |a_n|} =L|x|,$

assuming that $$\lim |a_{n+1}|/|a_n|$$ exists. Then the series converges if $$L|x| < 1$$, that is, if $$|x| < 1/L$$, and diverges if $$|x|>1/L$$. Only the two values $$x=\pm1/L$$ require further investigation. Thus the series will definitely define a function on the interval $$(-1/L,1/L)$$, and perhaps will extend to one or both endpoints as well. Two special cases deserve mention: if $$L=0$$ the limit is $$0$$ no matter what value $$x$$ takes, so the series converges for all $$x$$ and the function is defined for all real numbers. If $$L=\infty$$, then no matter what value $$x$$ takes the limit is infinite and the series converges only when $$x=0$$. The value $$1/L$$ is called the radius of convergence of the series, and the interval on which the series converges is the interval of convergence.

Consider again the geometric series, $$\sum_{n=0}^\infty x^n={1\over 1-x}.$$ Whatever benefits there might be in using the series form of this function are only available to us when $$x$$ is between $$-1$$ and $$1$$. Frequently we can address this shortcoming by modifying the power series slightly. Consider this series:

$\sum_{n=0}^\infty {(x+2)^n\over 3^n}= \sum_{n=0}^\infty \left({x+2\over 3}\right)^n={1\over 1-{x+2\over 3}}= {3\over 1-x},$

because this is just a geometric series with $$x$$ replaced by $$(x+2)/3$$. Multiplying both sides by $$1/3$$ gives

$\sum_{n=0}^\infty {(x+2)^n\over 3^{n+1}}={1\over 1-x},$

the same function as before. For what values of $$x$$ does this series converge? Since it is a geometric series, we know that it converges when

\eqalign{ |x+2|/3& < 1\cr |x+2|& < 3\cr -3 < x+2 & < 3\cr -5 < x& < 1.\cr }

So we have a series representation for $$1/(1-x)$$ that works on a larger interval than before, at the expense of a somewhat more complicated series. The endpoints of the interval of convergence now are $$-5$$ and $$1$$, but note that they can be more compactly described as $$-2\pm3$$. We say that $$3$$ is the radius of convergence, and we now say that the series is centered at $$-2$$.

Definition 11.8.3

A power series centered at $$a$$ has the form $$\sum_{n=0}^\infty a_n(x-a)^n,$$ with the understanding that $$a_n$$ may depend on $$n$$ but not on $$x$$.

## Contributors

• Integrated by Justin Marshall.