TITLE: Why do expressions of the form $\sum\limits^\infty_{n=1} \frac{n^k}{3^n}$ sum 'nicely'?
QUESTION [7 upvotes]: Why is it that sums like $\displaystyle\sum\limits^\infty_{n=1} \frac{n^k}{3^n}$ and $\displaystyle\sum\limits^\infty_{n=1} \frac{n^k}{2^n}$ where $k$ is a non-negative integer, sum to 'nice' numbers like 0.75, 1.5, 2, 4.125, 6, 15, 26 and 150 while sums like $\displaystyle\sum\limits^\infty_{n=1} \frac{n^k}{4^n}$ and $\displaystyle\sum\limits^\infty_{n=1} \frac{n^k}{7^n}$ don't?
More generally, how can one tell what a sum of the form $\displaystyle\sum\limits^\infty_{n=1} \frac{n^k}{m^n}$ sums to?

REPLY [2 votes]: Another way. Note that
$$
n^k=\sum_{m=0}^kS(k,m)n^{\underline{m}}
$$
where $S(k,m)$ is the stirling number of the second kind and $n^{\underline{m}}=n(n-1)\dotsb(n-m+1)$ is the falling factorial of length $m$. In effect we have simply changed bases to the more convenient basis of falling factorials which are a basis for polynomials of degree at most $k$. Now
$$
\sum_{n=0}^\infty n^k x^n=\sum_{n=0}^\infty\left(\sum_{m=0}^kS(k,m)n^{\underline{m}}\right)x^n
=\sum_{m=0}^kS(k,m)\left(\sum_{n=0}^\infty
n^{\underline{m}}x^n
\right)\quad (|x|<1)\tag{1}
$$
But it is easy to get the generating function of $n^{\underline{m}}$ by differentiation of the geometric series. Indeed,
$$
\sum_{n=0}^\infty n^{\underline{m}}x^n=x^m\left[\frac{d^m}{dx^m}(1-x)^{-1}\right]=\frac{m!x^m}{(1-x)^{m+1}}\quad (|x|<1)
$$
whence
$$
\sum_{n=0}^\infty n^k x^n=\sum_{m=0}^k \left[m!S(k,m)\frac{x^m}{(1-x)^{m+1}}\right].\quad(|x|<1).
$$
For example put $x=1/2$ to get that
$$
\sum_{n=0}^\infty n^k 2^{-n}=
2\sum_{m=0}^k m!S(k,m).
$$
and put $x=1/3$ to get that
$$
\sum_{n=0}^\infty n^k 3^{-n}=
3\sum_{m=0}^k m!S(k,m)\frac{1}{2^{m+1}}.
$$
and in general put $x=1/r$ where $|x|<1$, to get that
$$
\sum_{n=0}^\infty n^k r^{-n}=r\sum_{m=0}^km!S(k,m)\frac{1}{(r-1)^{m+1}}.
$$