Evaluating $\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$

Question

How can we evaluate:

$$\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$$

I tried to transform it into $$\sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$$$$=\ln \left(\sqrt{\frac{1}{\left(\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}\right)^2}+1}+\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$$ So that the original expression becomes: $$\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$$ $$=\ln\left(\prod_{n=1}^\infty\left(\sqrt{\frac{1}{\left(\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}\right)^2}+1}+\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)\right)$$ But it doesn't look good at all.
The final answer is close to 0.658479, which is $\ln\sqrt{2+\sqrt{3}}$

Answer 1

Note that $$\sinh^{-1} x-\sinh^{-1} y= \sinh^{-1} \left( x\sqrt{1+y^2}-y\sqrt{1+x^2} \right)$$

\begin{align*} \frac{1}{\sqrt{2^{n+2}+2}+\sqrt{2^{n+1}+2}} &= \frac{\sqrt{2^{n+2}+2}-\sqrt{2^{n+1}+2}} {(2^{n+2}+2)-(2^{n+1}+2)} \\ &= \frac{\sqrt{2^{n+2}+2}-\sqrt{2^{n+1}+2}} {2^{n+1}} \\ &= \sqrt{\frac{1}{2^{n}} \left( 1+\frac{1}{2^{n+1}} \right)}- \sqrt{\frac{1}{2^{n+1}} \left( 1+\frac{1}{2^{n}} \right)} \\ \sinh^{-1} \left( \frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}} \right) &= \sinh^{-1} \frac{1}{\sqrt{2^{n}}}-\sinh^{-1} \frac{1}{\sqrt{2^{n+1}}} \\ \sum_{n=1}^{\infty} \sinh^{-1} \left( \frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}} \right) &= \sinh^{-1} \frac{1}{\sqrt{2}} \\ &= \ln \left( \frac{1+\sqrt{3}}{\sqrt{2}} \right) \\ &= \ln \sqrt{2+\sqrt{3}} \end{align*}