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 $$\lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{n}\left(\frac{i}{n}\right)^{k} = \int_{0}^{1}x^{k}dx$$
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2$\begingroup$ Well, i'ts $1$ when $k=0$. $\endgroup$– MastremCommented Nov 9, 2015 at 9:09
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1$\begingroup$ Can't we use $$\lim_{n\to\infty}\sum_{i=1}^ni^k=\int_1^ni^kdi$$ ? $\endgroup$– GTX OCCommented Nov 9, 2015 at 9:10
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$\begingroup$ it's the sum of the area of all rectangles (under the curve $y = x^k$) with bases of length $1 \over n$ $\endgroup$– user49685Commented Nov 9, 2015 at 10:06
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$\begingroup$ @LutzL Should close the other one instead, this one is a bit more general. $\endgroup$– user99914Commented Nov 9, 2015 at 10:11
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$\begingroup$ 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. $\endgroup$– Lutz LehmannCommented Nov 9, 2015 at 10:13
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3 Answers
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$$\lim_{n\to\infty}\frac{1^{k}+2^{k}+...+n^{k}}{n^{k+1}}=\lim_{n\to\infty} \sum_{i=1}^{n} \frac{1}{n}\left(\frac{i}{n}\right)^{k} = \int_{0}^{1}x^{k}dx=\frac{1}{k+1}$$
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$\begingroup$ Can you elaborate how you made the limit an integral with those limits and $x^k$ please? $\endgroup$ Commented Nov 9, 2015 at 9:54
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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 $$ \lim \frac{1+2^{k}+...+n^{k}}{n^{k+1}}=\lim\frac{n^{k}}{n^{k+1}-(n-1)^{k+1}}\\ =\lim\frac{n^{k}}{(k+1)n^k-\frac{(k+1)k}2n^{k-1}+…}=\frac1{k+1} $$
answered Nov 9, 2015 at 9:56
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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.
answered Nov 9, 2015 at 9:17