Show that $\sum_{n=1}^\infty \frac{1}{\sqrt{n}(n+m)}\leq \frac{\pi}{\sqrt{m}}$

Question

Let $m\in\mathbb{N}^*$. Show that $$\sum_{n=1}^\infty \frac{1}{\sqrt{n}(n+m)}\leq \dfrac{\pi}{\sqrt{m}}$$

A hint is given by the problem: You could find this using an integration

How did I try to use the hint:

$$\sum_{n=1}^\infty \frac{1}{\sqrt{n}(n+m)}\leq \int_0^\infty\frac{1}{\sqrt{x}(x+m)}dx$$

If it's correct, I found this as a primitive function:

$2\sqrt m \arctan\left(\dfrac{\sqrt{x}}{\sqrt{m}}\right)$, that has the good limit as $x\rightarrow +\infty$.

Is that correct?

Answer 1

Hint: do you know any numerical methods for estimating an integral as a sum in some way? Can you cast the left hand side of your inequality as an approximation to an integral and then evaluate that integral?