Evaluate $\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \frac{k^p}{n^{p+1}}$

Question

Let $p$ be a real number. Evaluate $\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \dfrac{k^p}{n^{p+1}}$.

I think this depends on the value of $p$ because then large $n$ would mean small $n^{p+1}$ for small $p$ etc. Should I break this up into cases where $p > 1$ and $p \leq 1$? In any case, since the sum converges we have $$\lim_{n \to \infty} \sum_{k=1}^n \dfrac{k^p}{n^{p+1}} = \lim_{n \to \infty}\dfrac{1}{n^{p+1}} \sum_{k=1}^n k^p.$$ I am not sure how to continue since expanding $\displaystyle \sum_{k=1}^n k^p$ may be hard.

It looks like integrating $x^p$ from $0$ to $1$ to me. Hope this might help — k99731, Apr 26 '16 at 03:25

score 1 · Answer 1 · answered Apr 26 '16 at 03:31

1

HINT:

What is the Riemann sum for $\int_0^1 x^{p}\,dx$?

answered Apr 26 '16 at 03:31

Mark Viola

179,405

That is $\lim_{\Delta{x_i} \to 0}\sum_{k = 0}^{\infty}(x^*)^p(\Delta{x_i})$. How does that help? – user19405892 Apr 26 '16 at 03:35
Rewrite the limit as:
$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\frac{1}{n}\cdot \left( \frac{k}{n} \right )^p $$

Can you relate this to the formula for the Riemann sum you just posted?
– Fimpellizzeri Apr 26 '16 at 03:41
@Fimpellizieri I see the width length $\Delta_{x_i} = \dfrac{1}{n}$, but I don't see how to relate the other term. – user19405892 Apr 26 '16 at 03:52
@user19405892 The Riemann sum for $\int_0^1 x^p,dx$ is $\lim_{n\to \infty}\frac1n \sum_{k=1}^n (k/n)^p$. – Mark Viola Apr 26 '16 at 03:56
So the answer is $\frac{1}{2}$? – user19405892 Apr 26 '16 at 04:04
The answer is $\frac1{p+1}$. – Mark Viola Apr 26 '16 at 04:05

Evaluate $\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \frac{k^p}{n^{p+1}}$

1 Answers1