# Power Series

Recall that we were able to analyze all geometric series "simultaneously'' to discover that \[\sum_{n=0}^\infty kx^n = {k\over 1-x},\] 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 \(x\).

### Exercises 11.8

Find the radius and interval of convergence for each series. In exercises __3__ and __4__, do not attempt to determine whether the endpoints are in the interval of convergence.

**Ex 11.8.1** \(\sum_{n=0}^\infty n x^n\) (answer)

**Ex 11.8.2 **\(\sum_{n=0}^\infty {x^n\over n!}\) (answer)

**Ex 11.8.3 **\(\sum_{n=1}^\infty {n!\over n^n}x^n\) (answer)

**Ex 11.8.4** \(\sum_{n=1}^\infty {n!\over n^n}(x-2)^n\) (answer)

**Ex 11.8.5 **\(\sum_{n=1}^\infty {(n!)^2\over n^n}(x-2)^n\) (answer)

**Ex 11.8.6 **\(\sum_{n=1}^\infty {(x+5)^n\over n(n+1)}\) (answer)