What does $\sum_{n=1}^\infty\frac{1}{\sqrt n(n+1)}$ converge to exactly?
While I doubt there's a closed form, we can get a double integral easily enough:
$$\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{\pi}} \int_0^\infty e^{-n x^2} dx$$
$$\frac{1}{n+1}= \int_0^\infty e^{-(n+1) y} dy$$
Then the general term will be:
$$\frac{1}{\sqrt{n}(n+1)}=\frac{2}{\sqrt{\pi}} \int_0^\infty \int_0^\infty e^{-y} e^{-n (x^2+y)} ~dx ~dy$$
Let's try exchanging the summation and integration:
$$\sum_{n=1}^\infty e^{-n (x^2+y)}=\frac{1}{e^{x^2+y}-1}$$
Now we need the integral to converge, otherwise this whole thing would be in vain.
Numerically, using Mathematica, we have:
$$\frac{2}{\sqrt{\pi}} \int_0^\infty \int_0^\infty \frac{e^{-y}}{e^{x^2+y}-1} ~dx ~dy=1.8600250788 \dots$$
Which is quite in good agreement with the numerical value obtained for the series itself (the OEIS gives $1.86002507922119\dots$).
So, we can claim:
$$\sum_{n=1}^\infty\frac{1}{\sqrt n(n+1)}=\frac{2}{\sqrt{\pi}} \int_0^\infty \int_0^\infty \frac{e^{-y}}{e^{x^2+y}-1} ~dx ~dy$$
Why didn't I try to get a single integral? Because in that case after summation under the integral sign I got only divergent integrals.
Edit:
From the OEIS link we can actually see that the single integral also works. That's what I tried (only using my first integral and summing up the result with $1/(n+1)$, but I thought that the resulting integral of $$\int_0^\infty (-1-e^{x^2} \log (1-e^{-x^2}))dx$$ diverged. Apparently it doesn't.
Edit 2:
Let's try to make my double integral a little prettier.
$$\int_0^\infty \int_0^\infty \frac{e^{-y}}{e^{x^2+y}-1} ~dx ~dy=2\int_0^\infty \int_0^\infty \frac{z~e^{-z^2}}{e^{x^2+z^2}-1} ~dx ~dz=$$
$$x=r \cos t \\ z= r \sin t$$
$$=2\int_0^\infty \int_0^{\pi/2} \frac{r^2 \sin t~e^{-r^2 \sin^2 t}}{e^{r^2}-1} ~dt ~dr= \\ =2\int_0^\infty \int_0^1 \frac{r^2 e^{-r^2} e^{r^2 u^2}}{e^{r^2}-1} ~du ~dr=\sqrt{\pi} \int_0^\infty \frac{r ~\text{erfi}(r) e^{-r^2}}{e^{r^2}-1} ~dr$$
This integral converges very well and we get:
$$\sum_{n=1}^\infty\frac{1}{\sqrt n(n+1)}=2 \int_0^\infty \frac{r ~\text{erfi}(r) e^{-r^2}}{e^{r^2}-1} ~dr$$
Mathematica easily gives 50 correct digits of the series:
In[71]:= NIntegrate[
2 u Erfi[u] Exp[-u^2]/(Exp[u^2] - 1), {u, 0, Infinity},
WorkingPrecision -> 50]
Out[71]= 1.8600250792211903071806959157171433246665241215235
Or we can rewrite this in a slightly shorter form:
$$\sum_{n=1}^\infty\frac{1}{\sqrt n(n+1)}=\int_0^\infty \frac{\text{erfi}(\sqrt{r}) e^{-r}}{e^{r}-1} ~dr$$
We can also generalize the above to:
$$\sum_{n=1}^\infty\frac{1}{\sqrt n(n+a)}=\frac{1}{\sqrt{a}}\int_0^\infty \frac{\text{erfi}(\sqrt{ar}) e^{-ar}}{e^{r}-1} ~dr$$
Taking the limit for $a \to 0$, we recover the well known integral for the zeta function $\zeta(3/2)$.
Edit 3:
I'm sorry for bloating this post, but I wanted to make a few notes about numerical integration of the obtained expression.
It can be easily done by Gauss-Laguerre quadrature, provided we remove the singularity near $r=0$. Which can be done by getting the asymptotic expression as $r \to 0$:
$$\frac{\text{erfi}(\sqrt{r})}{e^{r}-1} \to \frac{2}{\sqrt{\pi r}}-\frac{\sqrt{r}}{3\sqrt{\pi}}+\frac{r^{3/2}}{30\sqrt{\pi}}+\frac{r^{5/2}}{315\sqrt{\pi}}+\dots$$
This singularity is exactly integrable and so we can write:
$$\int_0^\infty \frac{\text{erfi}(\sqrt{r}) }{e^{r}-1} e^{-r}~dr=2-\int_0^\infty \left(\frac{2}{\sqrt{\pi r}}-\frac{\text{erfi}(\sqrt{r}) }{e^{r}-1} \right) e^{-r}~dr$$
This allows us to approximate the integrand near $0$ by a well defined series:
$$\frac{2}{\sqrt{r}}-\sqrt{\pi} \frac{\text{erfi}(\sqrt{r})}{e^{r}-1}= \\ =\sqrt{r} \left(3-\frac{r}{30}-\frac{r^2}{315} +\frac{r^3}{1512}+\frac{r^4}{13860}-\frac{521 r^5}{32432400}-\frac{29 r^6}{16216200}+\dots \right) $$
The terms shown give the absolute error of the order $10^{-11}$ at $r=1/4$.
So we can denote:
$$f(r)=\frac{\sqrt{r}}{\sqrt{\pi}} \left(3-\frac{r}{30}-\frac{r^2}{315} +\frac{r^3}{1512}+\frac{r^4}{13860}-\frac{521 r^5}{32432400}-\frac{29 r^6}{16216200} \right) $$
And write:
$$\sum_{n=1}^\infty\frac{1}{\sqrt n(n+1)} \approx 2-\int_0^{1/4} f(r) e^{-r}~dr-\int_{1/4}^\infty \left(\frac{2}{\sqrt{\pi r}}-\frac{\text{erfi}(\sqrt{r}) }{e^{r}-1} \right) e^{-r}~dr$$
The first integral can taken exactly in terms of error function, while the second can be approximated by Gauss-Laguerre quadrature with very good accuracy.
Provided both the integrals above are computed with high enough precision, we get absolute error of the order $10^{-12}$ when compared to the numerical value for the series.
This also agrees with the bound from @robjohn's answer.
Not very convenient, but at least I have shown how to use this integral if one doesn't have advanced software like Mathematica.
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 $4\cdot 10^{-4}$.
Less than $\boldsymbol{2}$
Since $n+1\gt\frac12\left(\sqrt{n}+\sqrt{n+1}\right)\sqrt{n+1}$ , $$ \begin{align} \sum_{n=1}^\infty\frac1{\sqrt{n\vphantom{1}}\,(n+1)} &\lt\sum_{n=1}^\infty\frac2{\sqrt{n\vphantom{1}}\,\sqrt{n+1}\left(\sqrt{n\vphantom{1}}+\sqrt{n+1}\right)}\\ &=2\sum_{n=1}^\infty\left(\frac1{\sqrt{n\vphantom{1}}}-\frac1{\sqrt{n+1}}\right)\\[6pt] &=2\tag1 \end{align} $$
Numerical Value
The Euler-Maclaurin Sum Formula gives $$ \scriptsize\sum_{k=1}^n\frac1{\sqrt{k}\,(k+1)}=2\arctan\left(\sqrt{n}\right)+C-\pi+\frac1{2\sqrt{n}\,(n+1)}-\frac{3n+1}{24n^{3/2}(n+1)^2}+O\!\left(\frac1{n^{7/2}}\right)\tag2 $$ but we can handle this a bit easier if we expand the summand into a power series. $$ \frac1{\sqrt{n}\,(n+1)}=\sum_{k=0}^{11}\frac{(-1)^k}{n^{k+3/2}}+O\!\left(\frac1{n^{27/2}}\right)\tag3 $$ and apply the Euler-Maclaurin Sum Formula to $(3)$ to get $$ \begin{align} \scriptsize\sum_{k=1}^n\frac1{\sqrt{k}\,(k+1)} &\scriptsize=C-\frac2{n^{1/2}}+\frac7{6\,n^{3/2}}-\frac{41}{40\,n^{5/2}}+\frac{167}{168\,n^{7/2}}-\frac{1147}{1152\,n^{9/2}}\\ &\scriptsize+\frac{1411}{1408\,n^{11/2}}-\frac{200201}{199680\,n^{13/2}}+\frac{15331}{15360\,n^{15/2}}-\frac{1665833}{1671168\,n^{17/2}}\\[6pt] &\scriptsize+\frac{4371365}{4358144\,n^{19/2}}-\frac{83113379}{82575360\,n^{21/2}}+\frac{89747287}{90439680\,n^{23/2}}+O\!\left(\frac1{n^{25/2}}\right)\tag4 \end{align} $$ where $$ \scriptsize C=1.8600250792211903071806959157171433246665241215234514930491995\tag5 $$ $C$ was computed by summing the first $100000$ terms and using the series in $(4)$ evaluated at $n=100000$. Thus, $$ \sum_{k=1}^\infty\frac1{\sqrt{n}\,(n+1)}=C\tag6 $$