# $\frac{1}{n^s}-\int_n^{n+1} \frac{\mathrm{d}t}{t^s} \leq \frac{s}{n^{s+1}}$ looking for different proof

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?

• "Can we can find a sharper inequality?" You can solve the integral exactly. – Winther Oct 30 '18 at 15:01
• @Winther yes you are right... that was dumb on my part – auhasard Oct 30 '18 at 15:03
• 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. – Winther Oct 30 '18 at 15:07