Interesting log series Calculate the sum of series:
$$\sum_{k=1}^{\infty} k\left(\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-\log\left(\frac{k}{\sqrt{k^2+1}}\right)\right)$$
 A: We write
$$  k\left(\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-\log\left(\frac{k}{\sqrt{k^2+1}}\right)\right) = (k+1) \log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-k \log\left(\frac{k}{\sqrt{k^2+1}}\right) +\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right).$$
The sum of the first two terms is telescoping and we obtain that your sum $S$ is given by
$$S=  \overbrace{\lim_{k\to\infty} k \log\left(\frac{k}{\sqrt{k^2+1}}\right)}^{=0} - \overbrace{\lim_{k\to 1} k \log\left(\frac{k}{\sqrt{k^2+1}}\right)}^{=-\frac12 \ln2} - \overbrace{\sum_{k=1}^\infty \log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)}^{=S'}.$$
Now for $S'$, we use the rules of the logarithm, to obtain
$$S' = \lim_{n\to\infty} \sum_{k=1}^n \left(\log(k+1) -\frac12 \log\left((k+1)^2+1\right) \right).$$
The latter sum is evaluated  as $S'=\frac12 \log\left(\frac{2 \pi}{\sinh{\pi}}\right), see below.$
Thus, we obtain the final result
$$S= \frac{1}{2} \log\left(\frac{\sinh \pi}{\pi}\right).$$
Edit:
we are interested in the value of $$
S' = \lim_{n\to\infty} \sum_{k=1}^n \left(\ln(k+1) -\frac12 \log\left((k+1)^2+1\right) \right) =
\lim_{n\to\infty}  \left(\log \Gamma(n+2) - \frac12\sum_{k=1}^n\left[ \log\left(k+1+i\right) + \log\left(k+1-i\right)  \right] \right) 
= \lim_{n\to\infty} \left( \log\Gamma(n+2) - \frac12 \log\left[\Gamma(2+i +n)/\Gamma(2+i) \right]- \frac12 \log\left[\Gamma(2-i +n)/\Gamma(2-i) \right] \right) = \frac12  \left(\lim_{n\to\infty}
\log\left( \frac{\Gamma^2(n+2)}{\Gamma(2+i+n) \Gamma(2-i +n)} \right)  + \log[\Gamma(2+i)\Gamma(2-i) ]\right). $$
Now the ratio of the $\Gamma$ function in the first term approaches 1 and thus this term vanishes. Whereas in the second term, we use Euler's reflection formula to obtain the result quoted above.
