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 aswell 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 itsattractions: it appears to be an infinite version of one of thesimplest function types—a polynomial. This leads naturally to thequestions: Do other functions have representations as series? Is therean advantage to viewing them in this way?
The geometric series has a special feature that makes it unlike atypical polynomial—the coefficients of the powers of $x$ are thesame, namely $k$. We will need to allow more general coefficients ifwe are to get anything other than the geometric series.
Definition 11.8.1 A power series has the form $$\ds\sum_{n=0}^\infty a_nx^n,$$ with the understanding that $\ds a_n$ may depend on $n$ but not on$x$.$\square$
Example 11.8.2 $\ds\sum_{n=1}^\infty {x^n\over n}$ is a power series. We caninvestigate 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 theharmonic series and diverges; when $x=-1$ it is the alternatingharmonic series (actually the negative of the usual alternatingharmonic series) and converges. Thus, we may think of $\ds\sum_{n=1}^\infty {x^n\over n}$ as a function from the interval$[-1,1)$ to the real numbers.$\square$
A bit of thought reveals that the ratio test applied to a power serieswill 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 $\ds \lim |a_{n+1}|/|a_n|$ exists. Then the seriesconverges if $L|x|< 1$, that is, if $|x|< 1/L$, and diverges if$|x|>1/L$. Only the two values $x=\pm1/L$ require furtherinvestigation. Thus the series will definitely define a function onthe interval $(-1/L,1/L)$, and perhaps will extend to one or bothendpoints as well. Two special cases deserve mention: if $L=0$ thelimit is $0$ no matter what value $x$ takes, so the series convergesfor all $x$ and the function is defined for all real numbers. If$L=\infty$, then for any non-zero value of $x$ the limit is infinite,so the series converges only when $x=0$. The value $1/L$ is calledthe radius of convergence of the series, and theinterval 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 thisfunction are only available to us when $x$ is between $-1$ and$1$. Frequently we can address this shortcoming by modifying the powerseries 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 seriesconverge? Since it is a geometric series, we know that it convergeswhen $$\eqalign{ |x+2|/3&< 1\cr |x+2|&< 3\cr -3 < x+2 &< 3\cr -5< x&< 1.\cr}$$So we have a series representation for $1/(1-x)$ that works on alarger interval than before, at the expense of a somewhat morecomplicated series. The endpoints of the interval of convergence noware $-5$ and $1$, but note that they can be more compactly describedas $-2\pm3$. We say that $3$ is the radius of convergence, andwe now say that the series is centered at $-2$.
Definition 11.8.3 A power series centered at $a$ has the form$$\ds\sum_{n=0}^\infty a_n(x-a)^n,$$ with the understanding that $\ds a_n$ may depend on $n$ but not on$x$.$\square$
Exercises 11.8
Find the radius and interval of convergence for each series. Inexercises 3 and 4,do not attempt to determine whether the endpoints are in theinterval of convergence.
Ex 11.8.1$\ds\sum_{n=0}^\infty n x^n$(answer)
Ex 11.8.2$\ds\sum_{n=0}^\infty {x^n\over n!}$(answer)
Ex 11.8.3$\ds\sum_{n=1}^\infty {n!\over n^n}x^n$(answer)
Ex 11.8.4$\ds\sum_{n=1}^\infty {n!\over n^n}(x-2)^n$(answer)
Ex 11.8.5$\ds\sum_{n=1}^\infty {(n!)^2\over n^n}(x-2)^n$(answer)
Ex 11.8.6$\ds\sum_{n=1}^\infty {(x+5)^n\over n(n+1)}$(answer)
