Proof of $\lim_{n\to \infty}\frac 1 {n^{k+1}}\sum_{i=1}^n i^{k}=\frac 1 {k+1}$

Question

Prove the relation $\lim_{n\to \infty}\frac 1 {n^{k+1}}\sum_{i=1}^n i^{k}=\frac 1 {k+1}$ for any non negative integer k.

Do you know how to evaluate la sum ?. – Felix Marin Dec 15 '17 at 21:36 — Felix Marin, Dec 15 '17 at 21:36
Use Stolz-Ces$\mathrm{\grave{a}}$ro Theorem. – Felix Marin Dec 15 '17 at 21:57 — Felix Marin, Dec 15 '17 at 21:57

score 1 · Answer 1 · answered Dec 15 '17 at 21:41

1

$$\frac{(n-1)^{k+1}}{k+1}=\int_0^{n-1}x^kdx\le\sum_{j=1}^nj^k\le\int_1^nx^kdx=\frac{n^{k+1}-1}{k+1}$$

Now divide by $n^{k+1}$ and take limits.

answered Dec 15 '17 at 21:41

ajotatxe

score 1 · Accepted Answer · answered Dec 15 '17 at 21:42

1

Observe that

$$\frac {1}{n^{k+1}}\sum_{i=1}^ni^k=$$ $$\frac{1}{n}\sum_{i=1}^{n}(\frac{i}{n})^k $$ is a Riemann Sum of the function $x\mapsto x^k $.

its limit is the integral $$\int_0^1x^kdx=\frac {1}{k+1}-0$$

answered Dec 15 '17 at 21:42

hamam_Abdallah

2 Answers2