How does one work backwards from an infinite series to a function? How does one go about working backwards from an infinite series to a function?
For example, how would one determine what function this infinite series represents:
$$C + \frac{C}{x} + \frac{C}{x^2} + \frac{C}{x^3} +\cdots$$
where $C$ is any constant.
 A: It depends very much on the series. Your example, however, isn’t too hard, because it’s geometric:
$$\sum_{n\ge 0}\frac{C}{x^n}=C\sum_{n\ge 0}\left(\frac1x\right)^n=C\frac1{1-\frac1x}=\frac{Cx}{x-1}\;,$$
and it converges for $\left|\frac1x\right|<1$, i.e., for $|x|>1$.
A: Assuming $|x|>1$, this is a geometric series in $1/x$ and is
$$\frac{C}{1-(1/x)} = \frac{C x}{x-1}$$
A: Given a (not too crazy) function it is reasonably easy to derive a series; working backwards means recognizing it as some (variation on) a known series, like $\frac{1}{1 -y} = 1 + y + y^2 + \ldots$, which by $y \mapsto \frac{1}{x}$ leads to the series you give. The only way is to have a comprehensive list of known series, experience, and luck. A partial list of series you'll find when playing around with generating functions is given by Wilf in his "generatingfunctionology" in section 2.5.
A: A general method is to do something to the series that gets you the original back (vague, I know, but underscores multiple powerful ideas). Then solve for the series. In your case: If the series is $S$, then $S/x+C=S$. This means that $S=\frac{Cx}{x-1}$. In some cases you set up a differential equation:
$$S=\sum_{n=1}^\infty \frac{x^n}{n}$$
$$S'=\sum_{n=0}^\infty x^n$$
$$S'-1=xS'$$
$$S'=\frac{1}{1-x}$$
$$S=-\ln(1-x)$$
This doesn't handle every case but it can often help you analyze the function's behaviour, even if it doesn't lead to a closed form.
