Compute $\sum_{k=0}^{\infty} r^{k} k^{n}$

Question

Suppose $|r| < 1$ and $n$ is a nonnegative integer. Compute $\sum_{k=0}^{\infty}r^{k} k^{n}$.

I am asking this question because I know how to calculate $\sum_{k=0}^{\infty} r^{k} k(k-1)...(k - (n-1)) = \sum_{k=n}^{\infty} r^{k}k(k-1)...(k-(n-1))$ by pulling out derivatives. I am hoping that the method for doing this easy question can be applied to the series I want to calculate. I appreciate any help here. Thanks!

Edit: The best answer in (How can I evaluate $\sum_{n=0}^\infty(n+1)x^n$?) has an "added note" giving a general procedure for solving my problem, but it does not use power series. But every answer only fully answers my problem in the case where $n = 1$. I would like to see a more explicit answer for general $n$.

Answer 1

If you call $k^{(n)}:=k\cdot(k-1)\cdot\ldots\cdot\big(k-(n-1)\big)$, you can express any polynomial in $\mathbb R_{\le n}[k]$ (polynomials in $k$ of degree $n$ as much, together with zero polynomial) as a linear combination of $k^{(0)}=1,k^{(1)}=k,k^{(2)},\ldots,k^{(n)}$. Actually, there's exactly one such linear combination since $\{k^{(0)},k^{(1)},\ldots,k^{(n)}\}$ is a basis of the vector space $\mathbb R_{\le n}[k]$, as well as $\{k^0,k^1,\ldots,k^n\}$.

The coordinates in that basis, the coefficients of that linear combination, that is the numbers $a_i^n$ such that $$k^n=\sum_{i=0}^na_i^n k^{(i)},$$ are called Stirling numbers of the second kind and are easily calculated through the following recursion: $$a_0^0=1,$$ $$a_k^0=0,\quad k\ge1$$ $$a_0^n=0,\quad n\ge1$$ $$a_k^{n+1}=ka_k^n+a_{k-1}^n,\quad k\ge1,n\ge0.$$

So if you know how to calculate $\sum_{k=0}^\infty r^k k^{(n)}$ you just have to use the fact that $$\sum_{k=0}^\infty r^k k^n=\sum_{k=0}^\infty r^k \sum_{i=0}^na_i^n k^{(i)}=\sum_{i=0}^na_i^n \sum_{k=0}^\infty r^kk^{(i)},$$ which is just a linear combination of the sums $\sum_{k=0}^\infty r^k$, $\sum_{k=0}^\infty r^k k$, $\sum_{k=0}^\infty r^k k(k-1)$, etc.

Answer 2

$F_n(r) = \sum_{k=0}^\infty k^n r^k$ then $$F_{n+1} (r) = r \frac{\partial F_n}{\partial r}(r) $$

Now you can prove by induction that $$F_n (r) = \frac {P_n (r)}{(1-r)^{n+1}}$$ Whith $(P_n)_{n \ge 0}$ is a sequence of polynomials such that $$\left\{\begin{array}{ccl} P_0(X) &=& 1 \\ P_{n+1}(X) &=& X(1-X)P_n'(X) +(n+1)XP_n(X)\end{array}\right.$$

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Edit : Possible way to compute $P_n$ (recursivly)

You may observe that $\left(X^k(1-X)^{n-k}\right)_{k=0,\ldots,n}$ is a basis of $\mathbb R_n[X]$. We can write $$P_n(X) = \sum_{k=0}^n a_k^n X^k(1-X)^{n-k}$$

using the recursive relation, you will have $$a^{n+1}_k = \left\{\begin{array}{cl} 0 & k = 0 \\ ka_k^n + (2(n+1)-k)a^n_{k-1} & k=1,\ldots,n\\ (n+1)a_n^n & k=n+1 \end{array}\right.$$

Answer 3

The polylogarithm can be calculated as

$$ {\rm Li}_s(z) = \sum_{k=1}^{+\infty} \frac{z^k}{k^s} ~~\mbox{for}~~|z| < 1 $$

So your answer is

$$ \sum_{k=1}^{+\infty} r^k k^n = {\rm Li}_{-n}(r)~~\mbox{for}~~|r| < 1 $$

Answer 4

The Stirling Numbers of the Second Kind can be used here. One property they have is $$\newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}} \sum_{j=0}^k\stirtwo{n}{j}\binom{k}{j}j!=k^n\tag1 $$ Using $(1)$, we get $$ \begin{align} \sum_{k=0}^\infty r^kk^n &=\sum_{k=0}^\infty\sum_{j=0}^kr^k\stirtwo{n}{j}\binom{k}{j}j!\tag2\\ &=\sum_{j=0}^\infty\sum_{k=j}^\infty r^k\stirtwo{n}{j}(-1)^{k-j}\binom{-j-1}{k-j}j!\tag3\\ &=\sum_{j=0}^\infty\sum_{k=0}^\infty r^j(-r)^k\stirtwo{n}{j}\binom{-j-1}{k}j!\tag4\\ &=\sum_{j=0}^n\stirtwo{n}{j}j!\frac{r^j}{(1-r)^{j+1}}\tag5\\ &=\frac{\sum\limits_{j=0}^n\stirtwo{n}{j}j!(1-r)^{n-j}r^j}{(1-r)^{n+1}}\tag6 \end{align} $$ Explanation:
$(2)$: apply $(1)$
$(3)$: switch order of summation, use negative binomial coefficients
$(4)$: substitute $k\mapsto k+j$
$(5)$: apply Newton's Generalized Binomial Theorem
$(6)$: emphasize that the result is a polynomial in $r$, of degree at most $n$, divided by $(1-r)^{n+1}$