How can one prove the result of this limit?

Question

$$\lim _{n\to \infty }\left(\sum _{i=1}^n\:i^{\frac{1}{k}}\cdot \frac{1}{n^{1+\frac{1}{k}}}\right) = k/(1+k)$$ How can I prove the result of this limit, without resorting to integration?

Answer 1

Use the Stolz–Cesàro theorem with $a_n=\sum_{i=1}^ni^{1/k}$ and $b_n=n^{1+1/k}$. Then if you show that

$$\frac{a_{n}-a_{n-1}}{b_n-b_{n-1}}\to\frac k{k+1}$$

it would follow that

$$\frac{a_n}{b_n}=\frac1{n^{1+1/k}}\sum_{i=1}^ni^{1/k}\to\frac k{k+1}.$$

But

$$\frac{b_{n}-b_{n-1}}{a_n-a_{n-1}} =\underbrace{n\left(1-\left(1-\frac1n\right)^{1/k}\right)}_{\to 1/k} +\underbrace{\left(1-\frac1n\right)^{1/k}}_{\to1} \to\frac{k+1}k.$$

Answer 2

By Stolz-Cesaro

$$\lim _{n\to \infty } \sum _{i=1}^n\:i^{\frac{1}{k}}\cdot \frac{1}{n^{1+\frac{1}{k}}} =\lim _{n\to \infty } \frac{(n+1)^\frac1k}{ (n+1)^{1+\frac{1}{k}} -n^{1+\frac{1}{k}}} =\lim _{n\to \infty } \frac{n^\frac1k}{n^{1+\frac{1}{k}}}\frac{(1+1/n)^\frac1k}{ (1+1/n)^{1+\frac{1}{k}} -1 } =\lim _{n\to \infty } \frac{1}{n}\frac{ 1+o(1) }{ 1+\left(1+\frac1k\right)\frac1n+o(n^{-1})-1 } =\lim _{n\to \infty } \frac{ 1+o(1) }{ \left(1+\frac1k\right)+o(1)}=\frac{k}{k+1}$$