Geometric-power sums

[Inspired by Dominique Foata, Eulerian Polynomials: from Euler’s Time to the Present]

A quick note on: \[S_N^{(k)}(t)=\sum\limits_{n=1}^N n^k t^n\]

Using a similar trick to that used for summing geometric series, we look at: \[tS_N^{(k)}(t)=\sum\limits_{n=1}^N n^k t^{n+1}=\sum\limits_{n=2}^{N+1} (n-1)^k t^n\]

In fact, when \(k>0\) we can add include the \(n=1\) term in the sum since it is zero. Extracting the \(N+1\) term we then have: \[tS_N^{(k)}(t)=N^kt^{N+1}+ \sum\limits_{n=1}^N (n-1)^k t^n\]

Performing a binomial expansion on the \((k-1)^n\) term and extracting the sum we are interested in: \[tS_N^{(k)}(t)=N^kt^{N+1}+ S_N^{(k)}(t) + \sum\limits_{n=1}^N \sum\limits_{p=0}^{k-1}\binom{k}{p}(-1)^{k-p}n^p t^n\]

This creates a recursive means to evaluate these sums: \[tS_N^{(k)}(t)=N^kt^{N+1}+ S_N^{(k)}(t) + \sum\limits_{p=0}^{k-1}\binom{k}{p}(-1)^{k-p}S_N^{(p)}(t)\]

\[\implies S_N^{(k)}(t)=\dfrac{N^kt^{N+1}+ \sum\limits_{p=0}^{k-1}\binom{k}{p}(-1)^{k-p}S_N^{(p)}(t)}{t-1}\]

The recursion starts with the geometric series sum from \(n=1\): \[S_N^{(0)}=\dfrac{t^{N+1}-t}{t-1}=\sum\limits_{n=1}^N t^n\]