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.
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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
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1since $n$ is fixed, I don't think that Riemann Sums are really applicable – robjohn Sep 14 '15 at 19:56
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Oh I see. You guys are funny :) – user152169 Sep 14 '15 at 19:57
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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 $$ \lim_{p \to \infty} \frac{S_p(n)}{n^{p+1}} = \frac{1}{n}. $$
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1But 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
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1@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
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