Prove that $\lim \limits_{n \to ∞}\left(\frac{\sum_{i=1}^n k^p}{n^{p+1}}\right)$= $\left(\frac{1}{p+1}\right)$

Question

I know how to prove the case when $\lim \limits_{n \to ∞}$$\left(\frac{1} {n^{p}}\right)$ if p>0 but $\sum_{i=1}^n k^p$ it is causing me some problems , and the problem does not give me any clue.

http://math.stackexchange.com/questions/419765/limit-lim-n-to-infty-frac1p2p-ldotsnpnp1?lq=1 — Noam Shalev - nospoon, Sep 17 '15 at 20:20

score 2 · Answer 1 · answered Sep 17 '15 at 20:23

2

Rewrite the sum as $$\frac1n\biggl(\sum_{i=1}^n\Bigl(\frac kn\Bigr)^p\biggr)$$ One gets an upper Riemann sum for the function $x^p$.

answered Sep 17 '15 at 20:23

Bernard

175,478

Prove that $\lim \limits_{n \to ∞}\left(\frac{\sum_{i=1}^n k^p}{n^{p+1}}\right)$= $\left(\frac{1}{p+1}\right)$

1 Answers1