TITLE: What does $\sum_{n=1}^\infty\frac{1}{\sqrt n(n+1)}$ converge to exactly?
QUESTION [7 upvotes]: If
$$\sum_{n=1}^\infty\frac{1}{\sqrt n(n+1)}=S,$$
does $S$ have a known closed-form symbolic expression for its value? By the integral test it clearly converges, i.e. take $u=\sqrt x$ and $u$-substitute to get
$$\int_1^\infty\frac1{\sqrt x(x+1)}dx=2\arctan(\sqrt x)\rvert_1^\infty=\frac\pi2.$$
The sum was also in a competition problem, or so I'm told, to show that $S<2$ (if a source for this problem can be found, I'd appreciate it).

REPLY [4 votes]: To add still more less-than-closed forms, it can be also thought of in terms of integration of de Jonquiere's polylogarithm, $\mathrm{Li}_s(z)$. In particular, this function is defined by
$$\mathrm{Li}_s(z) := \sum_{n=1}^{\infty} \frac{z^n}{n^s}$$
Clearly one can see that
$$\int_{0}^{z} \mathrm{Li}_s(u)\ du = \sum_{n=1}^{\infty} \frac{z^{n+1}}{n^s (n+1)} = z \sum_{n=1}^{\infty} \frac{z^n}{n^s (n+1)}$$
and if you take $z = 1$ and $s = \frac{1}{2}$, you have the desired sum, that is,
$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}(n+1)} = \int_{0}^{1} \mathrm{Li}_\frac{1}{2}(t)\ dt$$
Sadly, the polylogarithm of fractional order does not integrate into anything else and Wolfram says it "has no expression in terms of standard mathematical functions".