Convergence of $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ I need to prove $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges. The root test is inconclusive, so I check in W.A., $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges by comparison test, but I don't know what series to compare. I've tried to compare with $\sum \frac{1}{4n+1} - \frac{1}{4n + 3}$ because this serie converges to $\frac{\pi}{4}$, but we have exactly the opposite inaquality $\frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} > \frac{1}{4n + 1} - \frac{1}{4n + 3}$.
Can you help me?

Hint $$\frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} = \frac{\sqrt{4n + 3}-\sqrt{4n + 1}}{\sqrt{4n + 3}\sqrt{4n + 1}}=\frac{2}{\sqrt{4n + 3}\sqrt{4n + 1}(\sqrt{4n + 3}+\sqrt{4n + 1})}.$$
