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I'm stuck in calculating the following limit: $$L=\displaystyle\lim_{n\to\infty}\dfrac{n(n+1)^\alpha}{\sum_{k=1}^nk^\alpha}$$ For what values of $\alpha\in\mathbb{R}$ $L$ has a finite value? Thanks.

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For $\alpha>-1$, by the Stolz–Cesàro theorem, \begin{eqnarray} L&=&\lim_{n\to\infty}\dfrac{n(n+1)^\alpha}{\sum_{k=1}^nk^\alpha}\\ &=&\lim_{n\to \infty}\frac{(n+1)(n+2)^\alpha-n(n+1)^\alpha}{(n+1)^\alpha}\\ &=&\lim_{n\to \infty}\frac{(n+1)(1+\frac{2\alpha}{n}+O(\frac{1}{n^2}))-n(1+\frac{\alpha}{n}+O(\frac{1}{n^2}))}{(1+\frac1n)^\alpha}\\ &=&\lim_{n\to \infty}\frac{1+\alpha+O(\frac{1}{n})}{(1+\frac1n)^\alpha}\\ &=&1+\alpha. \end{eqnarray} Here $$ (1+x)^\alpha=1+\alpha x+O(x^2). $$ For $\alpha<-1$, let $\beta=-\alpha$. Note that $\beta>1$ and $\{\sum_{k=1}^n\frac{1}{k^{\beta}}\}$ converges. Thus \begin{eqnarray} L&=&\lim_{n\to\infty}\dfrac{n(n+1)^{-\beta}}{\sum_{k=1}^nk^{-\beta}}\\ &=&\lim_{n\to \infty}\dfrac{n}{(n+1)^{\beta}\sum_{k=1}^n\frac{1}{k^{\beta}}}\\ &=&0. \end{eqnarray} For $\alpha=-1$, $\{\sum_{k=1}^n\frac{1}{k}\}$ diverges and hence \begin{eqnarray} L&=&\lim_{n\to\infty}\dfrac{n(n+1)^{-1}}{\sum_{k=1}^nk^{-1}}\\ &=&\lim_{n\to \infty}\dfrac{n}{(n+1)\sum_{k=1}^n\frac{1}{k}}\\ &=&0. \end{eqnarray}

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  • $\begingroup$ If $\alpha \le -1$, this development is invalid and the result is incorrect. $\endgroup$
    – Mark Viola
    Commented Jun 21, 2016 at 13:36
  • $\begingroup$ @Dr.MV, yes, you are right. Thanks. $\endgroup$
    – xpaul
    Commented Jun 21, 2016 at 13:38
  • $\begingroup$ @Dr.MV, the problem is fixed. $\endgroup$
    – xpaul
    Commented Jun 21, 2016 at 14:12
  • $\begingroup$ And a +1. Well done. -Mark $\endgroup$
    – Mark Viola
    Commented Jun 21, 2016 at 14:44

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