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?