Convergence of the following series. My professor asked me to prove that the series $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ is convergent and find the sum.
But I ended up proving it divergent. Here's my work :
Notice that $\sqrt{1+\frac{1}{n}}<\frac{4}{\sqrt{2}}-1$ for all positive integer $n$. This implies $\sqrt{2n}\times\sqrt{n+1}<4n-\sqrt{2}\times n$ for all positive integer $n$, i.e., $\sqrt{2n}\times(\sqrt{n}+\sqrt{n+1})<4n$ for all positive integer $n$.
Finally, we get $\frac{1}{4n}<\frac{1}{\sqrt{2n}(\sqrt{n}+\sqrt{n+1})}=\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ for all positive integer $n$.
Since $\sum_{n=1}^{\infty}\frac{1}{4n}$ is divergent, it follows that $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ is divergent by Comparison test.
Can someone tell me what's wrong with my proof ? Is the series really DIVERGENT ?

Yes the series diverges,
as $2n + 2 \geq n $ and $\sqrt{2n} \leq 2\sqrt{n}$,so $\sqrt{2n} + \sqrt{2n+2} \leq 2\sqrt{n} + \sqrt{n} = 3 \sqrt{n}$,
So,$\frac{1}{(\sqrt{2n}+\sqrt{2n+2})} \geq \frac{1}{3\sqrt{n}}$,
$\frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})} \geq \frac{1}{3n}$ or $\frac{1}{3n} \leq \frac{1}{\sqrt{n}(\sqrt{2n}+\sqrt{2n+2})}$ ,so by the comparision test the series diverges.
