I'm trying to find an expression for:

$$S(k) = \sum_{n=0}^{\infty} n^k \lambda^n$$ I have found a recursive expression for this where I first find $S(0)$, then use that to find $S(1)$. Then use those to find $S(2)$ and so on until I get to $S(k)$. For large $k$, I will need a computer for this. I want to know if there is a more efficient method, preferably a closed form solution I can calculate without a computer. Here is how my method works:

$$S(k) = \lambda + 2^k \lambda^2 + 3^k \lambda^3 + \dots$$ $$\frac{S(k)}{\lambda} = 1 + 2^k \lambda + 3^k \lambda^2 + \dots$$

Subtracting the first equation from the second equation we get:

$$S(k) \left(\frac{1-\lambda}{\lambda} \right) = (1^k-0^k)\lambda^0 + (2^k-1^k)\lambda^1+\dots$$ $$ = \sum_{n=0}^{\infty} ((n+1)^k-n^k)\lambda^n$$

$$=\sum_{n=0}^\infty \sum_{r=0}^{k-1} {k \choose r}n^r\lambda^n $$ $$ =\sum_{r=0}^{k-1}{k \choose r} \sum_{n=0}^\infty n^r \lambda^n$$ $$ = \sum_{r=0}^{k-1} {k \choose r} S(r)$$

So, $$S(k) = \frac{\lambda}{1-\lambda} \sum_{r=0}^{k-1} {k \choose r} S(r)$$

We know $S(0) = \frac{1}{1-\lambda}$. Can use this to get $S(1), S(2), \dots$. But is there a better way?

  • $\begingroup$ With $|\lambda|<1$ you get $S(k,\lambda) = \Phi(\lambda,-k,0)$ where $\Phi$ is the Lerch Transcendent $\endgroup$ – gammatester Oct 6 '18 at 7:59
  • $\begingroup$ This is related to the Eulerian polynomials (see Identities section), which in turn has a nice combinatoric interpretation in terms of descent numbers of permutations. $\endgroup$ – Sangchul Lee Oct 6 '18 at 8:00
  • $\begingroup$ This is the [polylogarithm function][1] $\operatorname{Li}_{-k}(\lambda)$. [1]: en.wikipedia.org/wiki/Polylogarithm $\endgroup$ – user10354138 Oct 6 '18 at 8:01
  • $\begingroup$ This is the polylogarithm of order $-kk. Google may help. $\endgroup$ – MPW Oct 6 '18 at 8:01
  • 1
    $\begingroup$ At least we have the following more explicit way of writing $S(k)$, which also appears in the wikipedia link above: $$ \sum_{i=0}^{\infty} i^k x^i = \frac{x \sum_{m=0}^{k-1} A(k,m)x^m}{(1-x)^{k+1}}, \qquad A(k,m)=\sum _{i=0}^{m}(-1)^{i}\binom{k+1}{i}(m+1-i)^k.$$ $\endgroup$ – Sangchul Lee Oct 6 '18 at 8:23

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