I want to show that $$ \sum_{k=1}^{n} k^{\alpha} \sim \frac{n^{\alpha+1}}{\alpha+1} \ \ \ \ \ \ n\to \infty$$

for $\alpha > -1$ . So I need to show that the following limit exists and is equal to $1$ : $$ \displaystyle \lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^{\alpha}}{\frac{n^{\alpha+1}}{\alpha+1}} $$

Because $\alpha > -1$, it is clear that the above limit is of the type $\frac{\infty}{\infty}$. Here I have doubts about using the L'Hospital's rule, because that means differentiating with respect to the upper bound of the sum which is a discrete variable, then how shoud I proceed with the computation of this limit?


  • $\begingroup$ use euler mac-laurin summation formula $\endgroup$ – tired Oct 18 '15 at 21:50
  • 1
    $\begingroup$ Or simply look at $$\frac{1}{n^{\alpha+1}}\sum_{k = 1}^n k^\alpha$$ and think Riemann sum. $\endgroup$ – Daniel Fischer Oct 18 '15 at 21:54

Use the inequality


which holds because the middle term is an upper sum for the left integral and a lower sum for the right integral. Equality holds if and only if $\alpha=0$. Now compute both integrals, divide throughout by ${n^{\alpha+1}}$ and let $n\to\infty$.

As pointed out by the8thone, the above holds for $\alpha\geq 0$, and for $-1<\alpha<0$ both inequalities are reversed, but the limits are still the same.

  • $\begingroup$ Thank you, however , the inequality you wrote is true for $\alpha \geq 0$, for $\alpha < 0$ we have the reverse inequalty, which finally gives the desired answer. $\endgroup$ – the8thone Oct 18 '15 at 22:16
  • $\begingroup$ You are absolutely correct. I'll fix it. $\endgroup$ – uniquesolution Oct 18 '15 at 22:16

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