How to compute $\sum_n(2n - \sqrt{n^2+1}-\sqrt{n^2-1})$? How to compute $\sum_n (2n - \sqrt{n^2+1}-\sqrt{n^2-1})$?
I tried two ways:
1. \begin{align*}
(2n - \sqrt{n^2+1}-\sqrt{n^2-1})
&= n - \sqrt{n^2+1} + n -\sqrt{n^2-1} \\
&= \frac{1}{n+\sqrt{n^2-1}}-\frac{1}{n-\sqrt{n^2+1}},
\end{align*}
but I don't know how to do later.
2. \begin{align*}
(2n - \sqrt{n^2+1}-\sqrt{n^2-1})
&= 2n - \frac{(\sqrt{n^2+1} + \sqrt{n^2-1})}{1} \\
&= 2n - \frac{2}{\sqrt{n^2+1} - \sqrt{n^2-1}},
\end{align*}
but I don't know how to do later too.
 A: Starting with Sangchul Lee's integral representation (which is a consequence of the Laplace transform)
$$ S = \int_{0}^{+\infty}\frac{I_1(x)-J_1(x)}{x(e^x-1)}\,dx = \frac{1}{\pi}\int_{0}^{+\infty}\int_{0}^{\pi}\frac{e^{-x\cos\theta}-e^{ix\cos\theta}}{e^x-1}\sin^2\theta\,d\theta\,dx \tag{1}$$
and applying Fubini's theorem we get
$$ S = \frac{1}{\pi}\int_{0}^{\pi}\left[\psi(1-i\cos\theta)-\psi(1+\cos\theta)\right]\sin^2\theta\,d\theta \tag{2}$$
where
$$\begin{eqnarray*} S &=& \frac{1}{2\pi}\int_{0}^{\pi}\left[\psi(1-i\cos\theta)+\psi(1+i\cos\theta)-2\psi(1-\cos\theta)\right]\sin^2\theta\,d\theta\\&=&\frac{1}{\pi}\int_{0}^{\pi}\left[\log\Gamma(1-\cos\theta)-\text{Im}\,\log\Gamma(1-i\cos\theta)\right]\cos\theta\,d\theta \tag{3}\end{eqnarray*}$$
allows an efficient numerical evaluation of $S$ through standard integration techniques (composite Simpson's rule or Gaussian quadrature):
$$ S \approx 0.6369740582412\tag{4} $$
but I do not believe that $S$ has a simple closed form in terms of standard mathematical constants.
We have a similar situation here.
A: Let us assume that you need to compute the infinite summation with
$$a_n=2n - \sqrt{n^2+1}-\sqrt{n^2-1}$$ For large values of $n$, rewrite $$a_n=n\left(2- \sqrt{1+\frac 1{n^2}}- \sqrt{1-\frac 1{n^2}}\right)$$ and use the binomial expansion or Taylor series to get
$$a_n=\frac{1}{4 n^3}+\frac{5}{64 n^7}+O\left(\frac{1}{n^{11}}\right)$$ So,
$$\sum_{n=1}^\infty a_n\approx\sum_{n=1}^p a_n+\frac 14 \sum_{n=p+1}^\infty \frac{1}{ n^3}=\sum_{n=1}^p a_n-\frac{1}{8}\psi ^{(2)}(p+1)\approx \sum_{n=1}^p a_n +\frac 1 {8 p^2}$$
If we use $p=10$, the first summation is $\approx 0.635843$, the correction term is $0.00125$ making a total of $0.637093$ while the infinite summation looks to be $\approx 0.636974$. For sure, if we increase $p$, we shall get closer and closer.
Now, the question is : what is this number ?
