Let $n \in \mathbb{N}_{>0}$ and $s \in \mathbb{R}_{>0}$ then I am interested in the following inequality :

$$\frac{1}{n^s}-\int_n^{n+1} \frac{\mathrm{d}t}{t^s} \leq \frac{s}{n^{s+1}}$$

Here is the way I prove this inequality :

Since $n \geq 1$ then the function : $t \mapsto \frac{1}{t^s}$ is dcreasing on $[n,n+1]$.Hence we have :

$$\frac{1}{n^s}-\int_n^{n+1} \frac{\mathrm{d}t}{t^s} \leq \frac{1}{n^s}-\frac{1}{(n+1)^s}$$

Now using the mean value theorem we get the desired inequality.

I am interested in other proof of this inequality, and if we can find a sharper inequality?

  • $\begingroup$ "Can we can find a sharper inequality?" You can solve the integral exactly. $\endgroup$ – Winther Oct 30 '18 at 15:01
  • $\begingroup$ @Winther yes you are right... that was dumb on my part $\endgroup$ – auhasard Oct 30 '18 at 15:03
  • 1
    $\begingroup$ As $n\to \infty$ the left hand side is $\sim \frac{s}{2n^{s+1}}$ so at best you can improve the inequality by a factor of $2$ on the right hand side. $\endgroup$ – Winther Oct 30 '18 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.