Convergent series and its sum Can anyone solve for the alternating sum of $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{\sqrt{1+n^2} 
}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{10}}\ ...$$ I know that the sum is convergent by the leibniz test but i cannot seem to find it.
 A: Since $\mathcal{L}^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)(s) = J_0(s)$ (a Bessel function of the first kind) the given series equals
$$\begin{eqnarray*} \int_{0}^{+\infty}\frac{J_0(s)}{e^s+1}\,ds &=& \sum_{n\geq 0}\frac{(-1)^n}{4^n n!^2}\int_{0}^{+\infty}\frac{s^{2n}}{e^s+1}\,ds\\&=&\color{blue}{\log 2+\sum_{n\geq 1}\left(-\frac{1}{16}\right)^n\binom{2n}{n} (4^n-1)\,\zeta(2n+1)}\\&=&\log 2+\frac{5\sqrt{2}-4\sqrt{5}}{10}+\sum_{n\geq 1}\left(-\frac{1}{16}\right)^n\binom{2n}{n} (4^n-1)\,\left(\zeta(2n+1)-1\right)\end{eqnarray*}$$
where the leftmost integral is simple to approximate numerically by exploiting $J_0(x)\approx \sqrt{\frac{2}{\pi x}}\cos\left(x-\frac{\pi}{4}\right)$ for $|x|\gg 1$ or
$$ \int_{0}^{+\infty}\frac{J_0(s)}{e^s+1}\,ds \approx \int_{0}^{+\infty}\frac{J_0(s)}{s+2}\,ds = \frac{\pi}{2}\left(H_0(2)-Y_0(2)\right) \approx 0.44.$$
A: Note that:
$$\sum_{n=0}^\infty\frac{(-1)^n}{\sqrt{1+n^2}}=\sum_{n=0}^\infty\frac1{(2n+i)^{1/2}(2n-i)^{1/2}}-\frac1{((2n+1)+i)^{1/2}((2n+1)-i)^{1/2}}$$
Thus, a closed form may be found in terms of the Shintani zeta function, not that I'd find this especially helpful.
