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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?

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

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Another possible solution: $$ 0 < a_n := \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} = \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4(n+1) - 1}} \\ < \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4(n+1) + 1}} =: b_n $$ and $\sum b_n$ is convergent as a telescoping series. It follows that $\sum a_n$ is convergent and $$ \sum_{n=0}^\infty a_n < \sum_{n=0}^\infty b_n = 1 \quad . $$

Or: The terms are positive, and the partial sums $$ \sum_{n=0}^N \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}} = 1 + \sum_{n=1}^N \left(-\frac{1}{\sqrt{4n - 1}} + \frac{1}{\sqrt{4n + 1}} \right) - \frac{1}{\sqrt{4N + 3}} < 1 $$ are bounded. (This is essentially what achille hui suggested in the above comment.)

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Asmptotics.
As $n \to +\infty$: $$\begin{align} (4n+1)^{-1/2} &= (4n)^{-1/2}\left(1+\frac{1}{4n}\right)^{-1/2} =\frac{1}{2\sqrt{n}}\left(1-\frac{1}{8n}+O(n^{-2})\right) \\ (4n+3)^{-1/2} &= (4n)^{-1/2}\left(1+\frac{3}{4n}\right)^{-1/2} =\frac{1}{2\sqrt{n}}\left(1-\frac{3}{8n}+O(n^{-2})\right) \\ (4n+1)^{-1/2} - (4n+3)^{-1/2}&= \frac{1}{2\sqrt{n}}\left(0+\frac{2}{8n}+O(n^{-2})\right) =\frac{1}{8n^{3/2}}+O(n^{-5/2}) \end{align}$$ The series converges by comparison with $$\sum\frac{1}{8n^{3/2}}$$

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