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The series $\sum\limits_{k=1}^{\infty}\frac{1}{\sqrt{2k+\sqrt{4k^2-1}}}$ is divergent. I am interested in its partial sums to do some computations based on them.

I tried to multiply $\sqrt{2k+\sqrt{4k^2-1}}$ by $\sqrt{2k-\sqrt{4k^2-1}}$ and to divide by the same term, but I have got no telescoping sum for evaluation. Maybe I should use others technics.

Now my question is what is the closed form of the sum $$\sum_{k=1}^{n}\frac{1}{\sqrt{2k+\sqrt{4k^2-1}}}\ ?$$

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Using the identity

$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{1}{2} \left(a+\sqrt{a^2-b}\right)}+\sqrt{\frac{1}{2} \left(a-\sqrt{a^2-b}\right)}$$

where: $a = 2 k$ and $b=4 k^2-1$, along with evaluating a telescoping series, we find that

$$\begin{align} \color{red}{\sum _{k=1}^n \frac{1}{\sqrt{2 k+\sqrt{4 k^2-1}}}}&=\sum _{k=1}^n \frac{\sqrt{2}}{\sqrt{-1+2 k}+\sqrt{1+2 k}}\\\\ &=\sum _{k=1}^n \frac{\sqrt{2} \left(\sqrt{2 k+1}-\sqrt{2 k-1}\right)}{\left(\sqrt{-1+2 k}+\sqrt{1+2 k}\right) \left(\sqrt{2 k+1}-\sqrt{2 k-1}\right)}\\\\ &=\frac{\sum _{k=1}^n \left(\sqrt{2 k+1}-\sqrt{2 k-1}\right)}{\sqrt{2}}\\\\ &=\frac{-1+\sqrt{1+2 n}}{\sqrt{2}}\\\\ &=\color{red}{-\frac{1}{\sqrt{2}}+\frac{\sqrt{1+2 n}}{\sqrt{2}}} \end{align}$$

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  • $\begingroup$ Thanks so much for the slick answer $\endgroup$ – zeraoulia rafik Jul 22 '18 at 17:09
  • $\begingroup$ @zeraouliarafik . You're welcome :) $\endgroup$ – Mariusz Iwaniuk Jul 22 '18 at 17:20
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    $\begingroup$ (+1) Nicely done. But CAS is unnecessary. Simply multiply numerator and denominator by $\sqrt(2k+1)-\sqrt(2k-1)$ to yield a telescoping series. $\endgroup$ – Mark Viola Jul 22 '18 at 18:18
  • $\begingroup$ @MarkViola. Thanks for HINT. $\endgroup$ – Mariusz Iwaniuk Jul 22 '18 at 18:40
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Jul 23 '18 at 21:43

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