On asymptotic properties of the sum of consecutive powers

Question

For positive integers $p, n$, the sum of consecutive $p$-th powers is $$ S_p(n) := \sum_{k=1}^n k^p. $$ From Asymptotic behaviour of sums of consecutive powers we have that $S_p(n)/n^{p+1}$ is decreasing when viewed as a function of $p$ (with $n$ fixed). Can we be more precise? Say can we compute the limit of $S_p(n)/n^{p+1}$ as $p \to \infty$? Thanks.

Answer 1

We have $$ \frac{S_p(n)}{n^{p+1}} = \frac{1}{n} \left( \frac{1}{n^p} + \left(\frac{2}{n}\right)^p + \dotsb + \left(\frac{n-1}{n}\right)^p + 1 \right). $$ Now, $n$ is fixed, so if we take the limit as $p \to \infty$, we can push it inside the sum. Every term in the sum is the $p$th power of a number in $(0,1)$ apart from the last one. Therefore every term in the sum tends to zero except the last one, and $$ \lim_{p \to \infty} \frac{S_p(n)}{n^{p+1}} = \frac{1}{n}. $$