# On asymptotic properties of the sum of consecutive powers

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.

• I am not following: since $n$ is fixed, it suffices to compute the limit of $k^p/n^{p+1}$ for each $1\le k\le n$. Is this out of reach? – Did Sep 14 '15 at 19:54
• since $n$ is fixed, I don't think that Riemann Sums are really applicable – robjohn Sep 14 '15 at 19:56
• Oh I see. You guys are funny :) – user152169 Sep 14 '15 at 19:57

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 $$\lim_{p \to \infty} \frac{S_p(n)}{n^{p+1}} = \frac{1}{n}.$$
• But the number of terms in the sum is also a function of $p$. Something needs to be said to justify interchanging the sum and limit. – Jair Taylor Sep 14 '15 at 19:37
• @JairTaylor: no, $n$ is fixed. The number of terms does not change. I was about to post the same answer (with some notation changes), so I am pretty sure that this is correct. – robjohn Sep 14 '15 at 19:49