Limit of sum $\sum_{x=1}^{\infty} \frac{1}{\sqrt{x}\sqrt{x+1}}$ I am having a hard time proving if this sum converges or diverges:
$$\sum_{x=1}^{\infty} \frac{1}{\sqrt{x}\sqrt{x+1}}$$
I tried proving it by the ratio test but $q = 1$.
I couldn’t proceed further and would like some help.

Dunno if it helps, but $\sqrt x<\sqrt {x+1}$ because the square root is increasing, and therefore
$$\frac1{\sqrt x\sqrt{x+1}}>\frac1{\sqrt {x+1}\sqrt {x+1}}=\frac1{x+1}$$
because $x\mapsto 1/x$ is decreasing for positive $x$.  Thus you are left with the harmonic series (minus 1).
