TITLE: Evaluating $\sum _{n=1}^{\infty } \sinh ^{-1}\left(\frac{1}{\sqrt{2^{n+1}+2}+\sqrt{2^{n+2}+2}}\right)$
QUESTION [5 upvotes]: 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 $\ln\sqrt{2+\sqrt{3}}$

REPLY [4 votes]: 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*}