Closed-form expression of $\frac{1}{N^{n}} \left[ N^{n+1} - \sum_{k=1}^{N} (k-1)^{n} \right] $

Question

Is there a nice closed-form expression for $$\frac{1}{N^{n}} \left[ N^{n+1} - \sum_{k=1}^{N} (k-1)^{n} \right] $$

where $n, N, k \in \mathbb{N}$.

I can obtain an approximation for this for large $N$, since if $N$ is large then we can approximate it by $$\int_{0}^{N} y^{n}dy = \frac{N^{n+1}}{n+1}$$

so that the original expression is approximately equal to $$ \frac{nN}{n+1} $$

But is there a way to represent my quantity without resorting to approximations?

Answer 1

For the sum

$$ \sum_{k=1}^{N} (k-1)^{n} = \sum_{k=0}^{N-1} k^{n} $$

you can use the well known formula in terms of the Bernoulli numbers

$$ \sum_{i=0}^n i^p = \frac{(n+1)^{p+1}}{p+1} + \sum_{k=1}^p\frac{B_k}{p-k+1}{p\choose k}(n+1)^{p-k+1}, $$