2

$$\lim_{n\rightarrow \infty}n\left[\frac{1}{n^{\alpha +1}}(1^{\alpha}+2^{\alpha}+\cdots+n^{\alpha})-\frac{1}{\alpha+1}\right]=\frac{1}{2}$$

I know that this can be solved by the Stolz theorem, as there, but when I do it myself, I first try this: $$\text{The origen}=\frac{\frac{1}{n}\sum_{i=1}^{n}(\frac{i}{n})^\alpha-\frac{1}{\alpha+1}}{\frac{1}{n}}$$, it seems that the summation is somehow like $\int_{0}^{1}x^{\alpha}dx$, but I do not know what to do next, Could it can be done following my method? Thank you!

StubbornAtom
  • 17,052
fractal
  • 551

1 Answers1

1

An appproach using Faulhaber formula.

Writing the limit $$L= \lim_{n\to \infty} \left(\frac{1}{n^{\alpha}}\sum_{k=1}^n k^{\alpha }-\frac{n}{\alpha +1}\right)$$ Using the Faulhaber formula we have the limit as $$\begin{aligned}L=\lim_{n\to\infty}\left(\frac{1}{n^{\alpha} }\left(\frac{n^{\alpha +1}}{\alpha +1} +\frac{n^{\alpha}}{2}+\sum_{k=2}^{\alpha} \frac{B_k}{k!}(\alpha)_{k-1} n^{\alpha -k+1}\right)-\frac{n}{\alpha+1}\right)=\lim_{n\to\infty}\left(\frac{1}{2}+\frac{1}{n^{k-1}}\sum_{k=1}^{\alpha}\frac{B_k}{k!}\frac{\alpha!}{(\alpha -k+1)!}\right)=\frac{1}{2}+\lim_{n\to \infty}\frac{F(\alpha,k)}{n^{k-1}}=\frac{1}{2} +0=\frac{1}{2}\end{aligned}$$ Here $B_k$ is kth Bernoulli number and$\displaystyle (\alpha)_{k-1} =\frac{\alpha !}{(\alpha-k+1)!}$ and since we are interested in evaluating the limit so the limit of latter sum $F(\alpha,k)$ is clear to us which is $0$.

Naren
  • 3,432