What does $\sum_{n=1}^\infty\frac{1}{\sqrt n(n+1)}$ converge to exactly? 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).
 A: This adds pretty little to the previous answer: the (inverse) Laplace transform allows to write the given series as the integral $\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{F(\sqrt{s})}{e^s-1}\,ds$ where $F$ is Dawson's function. $F(\sqrt{s})$ approximately behaves like $\sqrt{s} e^{-2s/3}$ and any algorithm for numerical integration is able to provide a decent approximation for the mentioned integral.
This actually brings something new to the table: by hand, me may realize that
$$ \color{red}{S}=\sum_{n\geq 1}\frac{1}{(n+1)\sqrt{n}} \color{red}{\approx} \frac{1}{2}+\sqrt{\pi}\sum_{n\geq 2}\frac{\binom{2n}{n}}{4^n(n+1)}=\color{red}{\frac{1}{2}+\frac{3}{4}\sqrt{\pi}}$$
since $\frac{1}{4^n}\binom{2n}{n}\approx\frac{1}{\sqrt{\pi n}}$ is a  pretty good approximation for any $n\geq 1$ and the generating function for Catalan numbers is fairly well-known. This can be improved by exploting the more accurate
$$ \frac{1}{\sqrt{n}}\approx\frac{\sqrt{\pi}}{4^n}\binom{2n}{n}\left(1+\tfrac{1}{8n}+\tfrac{1}{128n(n+2)}\right).$$
Creative telescoping also deserves a try: indeed, $\frac{1}{(n+1)\sqrt{n}}<\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}$ immediately proves $\color{red}{S<2}$, and the more accurate $\frac{1}{(n+1)\sqrt{n}}\approx \frac{2}{\sqrt{n+\frac{1}{6}}}-\frac{2}{\sqrt{n+\frac{7}{6}}}$ gives $\color{red}{S\approx \frac{1}{2}+2\sqrt{\frac{6}{13}}}$.

We may also combine the approximation through central binomial coefficients with the Cauchy-Schwarz inequality to get an exceptionally simple and very accurate approximation:
  $$ S\leq \frac{1}{2}+\sqrt{\left(\sum_{n\geq 2}\frac{4^n}{n(n+1)\binom{2n}{n}}\right)\left(\sum_{n\geq 2}\frac{\binom{2n}{n}}{(n+1)4^n}\right)}=\frac{1}{2}+\sqrt{\frac{\pi^2}{4}\cdot\frac{3}{4}}$$
  gives $\color{red}{S\approx \frac{1}{2}+\frac{\sqrt{3}}{4}\pi}$, whose absolute error is less than $4\cdot 10^{-4}$.

A: 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".
