# Evaluate $\lim_{n\to\infty}\frac{1^{k}+2^{k}+...+n^{k}}{n^{k+1}}$ [duplicate]

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

• Well, i'ts $1$ when $k=0$. Commented Nov 9, 2015 at 9:09
• Can't we use $$\lim_{n\to\infty}\sum_{i=1}^ni^k=\int_1^ni^kdi$$ ? Commented Nov 9, 2015 at 9:10
• it's the sum of the area of all rectangles (under the curve $y = x^k$) with bases of length $1 \over n$ Commented Nov 9, 2015 at 10:06
• @LutzL Should close the other one instead, this one is a bit more general.
– user99914
Commented Nov 9, 2015 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. Commented Nov 9, 2015 at 10:13

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

• Can you elaborate how you made the limit an integral with those limits and $x^k$ please? Commented Nov 9, 2015 at 9:54
• Nice answer ... +1 Commented Nov 9, 2015 at 9:55

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 $$\lim \frac{1+2^{k}+...+n^{k}}{n^{k+1}}=\lim\frac{n^{k}}{n^{k+1}-(n-1)^{k+1}}\\ =\lim\frac{n^{k}}{(k+1)n^k-\frac{(k+1)k}2n^{k-1}+…}=\frac1{k+1}$$

• Nice one, and refreshing.. Commented Nov 9, 2015 at 10:36

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.