Closed form of $\sum_{i=0}^{\infty} x^i (i!(i+1)!)^{-1/2}$ I am looking for a closed form of this series
$$
\sum\limits_{i=0}^{\infty} \frac{x^i}{\sqrt{i!(i+1)!}},\:x \geq 0.
$$
The main problem is that this series contains roots of factorials. But many special functions just contain factorials, not their roots. Is it possible to express the sum of this series in some well-known (or little known) functions?

You will not find closed form, but using
$$n^{-s} = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} e^{-nt} \, {\rm d}t$$
for $s=1/2$ (it should work for general $s$) and substituting $e^{-t} \rightarrow t$, you can bound your sum above $$f(x)=\sum_{n=0}^\infty \frac{x^n}{n!\sqrt{n+1}} = \frac{1}{\sqrt{\pi}}\int_0^1 \frac{e^{xt}}{\sqrt{-\log(t)}} \, {\rm d}t \stackrel{t=1-\tau}{=} \frac{e^x}{\sqrt{\pi}}\int_0^1 \frac{e^{-x\tau}}{\sqrt{-\log(1-\tau)}} \, {\rm d}\tau \\ \leq \frac{e^x}{\sqrt{\pi}}\int_0^1 \frac{e^{-x\tau}}{\sqrt{\tau}} \, {\rm d}\tau \stackrel{x\tau=u^2}{=} \frac{e^x}{\sqrt{x}} \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{x}} e^{-u^2} \, {\rm d}u \leq \frac{e^x}{\sqrt{x}}\, ,$$
where $-\log(1-t) \geq t$ was used.
You can find better approximations, by expanding
$$\frac{1}{\sqrt{-\ln(1-t)}} = \frac{1}{\sqrt{t}} - \frac{\sqrt{t}}{4} - \frac{7t^{3/2}}{96} + O\left(t^{5/2}\right)$$
about $t=0$, and which leads to
$$f(x)= \frac{e^x}{\sqrt{x}} \left( 1 - \frac{1}{8x} - \frac{7}{128x^2} + O\left(1/x^3\right) \right) {\rm erf}\left( \sqrt{x} \right) + \frac{1}{\sqrt{\pi}} \left( \frac{31}{96x} + \frac{7}{64x^2} \right) \\
= \frac{e^x}{\sqrt{x}} \left( 1 - \frac{1}{8x} - \frac{7}{128x^2} + O\left(1/x^3\right) \right) \, .$$
Furthermore I played around via try and error an looked for smooth elementary functions that match $f(x)=1+\frac{x}{\sqrt{2}}+O(x^2)$ up to first order about $x=0$ and till second order asymptotically. My ansatz was
$$f(x)=e^x \sqrt{\frac{1+c/x+1/x^2}{x+a+b/x+1/x^2}}$$
and Taylor expanding about $x=0$ and $x=\infty$ gives the coefficients
$$a=\frac{4\sqrt{2}}{3}-\frac{7}{8} \\
b=\frac{\sqrt{2}}{3} + \frac{7}{8} \\
c=\frac{4\sqrt{2}}{3} - \frac{9}{8}$$
which approximates $f(x)$ reasonably well and bounds it from above. You can do much better by increasing the order inside the square-root from $1/x^2$ to $1/x^3$ and using the two new degrees of freedom to match up to third order about $x=0$. The expressions for the coefficients become more complicated though.
Here is a graph:

