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I need to find $$\lim_{n\to\infty}\frac{1^{k}+2^{k}+...+n^{k}}{n^{k+1}}$$ for $k>0$ And I just need an explanation on how come $$\lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{n}\left(\frac{i}{n}\right)^{k} = \int_{0}^{1}x^{k}dx$$

user114138
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    Well, i'ts $1$ when $k=0$. – Mastrem Nov 09 '15 at 09:09
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    Can't we use $$\lim_{n\to\infty}\sum_{i=1}^ni^k=\int_1^ni^kdi$$ ? – GTX OC Nov 09 '15 at 09:10
  • it's the sum of the area of all rectangles (under the curve $y = x^k$) with bases of length $1 \over n$ – user49685 Nov 09 '15 at 10:06
  • @LutzL Should close the other one instead, this one is a bit more general. –  Nov 09 '15 at 10:11
  • I was not sure what the rules of chronological precedence are. Perhaps one should find an identical question, there was one in the last 3 days. – Lutz Lehmann Nov 09 '15 at 10:13
  • Earlier duplicates: https://math.stackexchange.com/q/150391/115115, https://math.stackexchange.com/q/936924/115115, https://math.stackexchange.com/q/478344/115115. – Lutz Lehmann Nov 09 '15 at 10:17
  • To the OP: Try to write down $\int_0^1 x^k dx$ as Riemann sum. –  Nov 09 '15 at 12:41
  • @LutzL : Now there are so many duplicates, I am not so sure which one to close now. I've flagged for moderator attention. We will see. –  Nov 09 '15 at 12:45

3 Answers3

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$$\lim_{n\to\infty}\frac{1^{k}+2^{k}+...+n^{k}}{n^{k+1}}=\lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{n}\left(\frac{i}{n}\right)^{k} = \int_{0}^{1}x^{k}dx=\frac{1}{k+1}$$

user2280549
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From https://math.stackexchange.com/a/1138452/115115

Via Cesaro-Stolz, the discrete version of l'Hopital: $$ \lim\frac{a_1+...+a_n}{b_1+...+b_n}=\lim \frac{a_n}{b_n} $$ under suitable assumptions, essentially that the second limit exists, we get here $$ \lim \frac{1+2^{k}+...+n^{k}}{n^{k+1}}=\lim\frac{n^{k}}{n^{k+1}-(n-1)^{k+1}}\\ =\lim\frac{n^{k}}{(k+1)n^k-\frac{(k+1)k}2n^{k-1}+…}=\frac1{k+1} $$

Lutz Lehmann
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May be you already know that the numerator is a generalized harmonic number $$\sum_{i=1}^n i^k=H_n^{(-k)}$$ An asymptotic expansion is $$H_n^{(-k)}=n^k \left(\frac{n}{k+1}+\frac{1}{2}+\frac{k}{12 n}+O\left(\left(\frac{1}{n}\right)^3\right)\right)+\zeta (-k)$$ This makes $$\frac{\sum_{i=1}^n i^k}{n^{k+1}}=\frac{1}{k+1}+\frac{1}{2 n}+\cdots$$ It is more complex than user2280549's clear and simple answer; I wrote my naswer to show how is approached the limit.

Claude Leibovici
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