11.9: Power Series

Recall that we were able to analyze all geometric series "simultaneously'' to discover that $$\sum_{n=0}^\infty kx^n = {k\over 1-x}, \nonumber$$ 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 $$\sum_{n=0}^\infty kx^n \nonumber$$ 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:

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

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

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

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

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

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:

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

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

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

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

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

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

David Guichard (Whitman College)

• Integrated by Justin Marshall.