Find a limit for x.

Question

Find a limit for the following sequence: $$x_n = \frac{1^p+2^p+...+n^p}{n^{p+1}} ; p\ge{1}$$ Can you also explain me how to solve it? I tried many times but I didn't manage to do it.

Answer 1

By Faulhaber's formula, $1^p + \cdots + n^p = \dfrac1{p+1} n^{p+1} + \dfrac12 n^p + O(n^{p-1})$, so the limit is $\dfrac1{p+1}$.

Answer 2

It's a particular case of certain integral: $$\lim_{n\to\infty} x_n = \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^n (\frac{k}{n})^p = \int_0^1 x^pdx = \frac{x^{p+1}}{p+1}\big|_0^1 = \frac{1}{p+1} $$

Answer 3

Hint:

Rewrite this sum as $$x_n=\frac1n\sum_{k=1}^n\Bigl(\frac kn\Bigr)^p.$$ Can you recognise a Riemann sum?

Answer 4

Let $\frac{1}{n}=x$.

Thus, by Stolz and L'hospital we obtain: $$\lim_{n\rightarrow+\infty}x_n=\lim_{n\rightarrow+\infty}\frac{n^p}{n^{p+1}-(n-1)^{p+1}}=$$ $$=\lim_{x\rightarrow0}\frac{x}{1-(1-x)^{p+1}}=\lim_{x\rightarrow0}\frac{1}{(p+1)(1-x)^p}=\frac{1}{p+1}.$$

Answer 5

By Stolz-Cesaro we have

$$\lim_{n\to \infty}\frac{1^p+2^p+...+n^p}{n^{p+1}}=\lim_{n\to \infty}\frac{(n+1)^p}{(n+1)^{p+1}-n^{p+1}}=\lim_{n\to \infty}\frac{(1+\frac1n)^p}{(n+1)(1+\frac1n)^{p}-n}=\frac1{1+p}$$

indeed

$$(1+\frac1n)^p=1+\frac p n+o(1/n)$$

and therefore

$$\frac{(1+\frac1n)^p}{(n+1)(1+\frac1n)^{p}-n}=\frac{(1+\frac1n)^p}{(n+1)(1+\frac p n+o(1/n))-n}=\frac{(1+\frac1n)^p}{n+p+1+\frac pn-n+o(1)}=$$

$$\frac{(1+\frac1n)^p}{p+1+\frac pn+o(1)}\to\frac1{1+p}$$