How can I prove that $\lim_{n \to \infty} \frac{1^p+2^p+\cdots+n^p}{n^{p+1}} = \int_{0}^{1} x^p dx$?

Question

I have to prove that $\lim_{n \to \infty} \frac{1^p+2^p+\cdots+n^p}{n^{p+1}} = \int_{0}^{1} x^p dx.$

I tried to calculate the limit, but I'm going to nowhere. I could prove the first part equals to $\frac{1}{p+1}$ with Stolz-Cesàro and resolved the integral, but can't relate the two of them besides that. Is it only luck or is there a real relation between the two?

See http://math.stackexchange.com/questions/465075/find-lim-limits-n-to-infty-frac1n-sum-limits2n-r-1-fracr-sq OR http://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2 — lab bhattacharjee, Sep 17 '15 at 15:10
Do you know that a definite integral can be written as an infinite sum? — najayaz, Sep 17 '15 at 15:17

score 4 · Answer 1 · answered Sep 17 '15 at 15:17

4

$x^p$ is a continuous function over $[0,1]$, hence a Riemann-integrable one. That gives:

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

answered Sep 17 '15 at 15:17

Jack D'Aurizio

353,855

How can I prove that $\lim_{n \to \infty} \frac{1^p+2^p+\cdots+n^p}{n^{p+1}} = \int_{0}^{1} x^p dx$?

1 Answers1