Closed form of $\sum_{i=0}^{\infty} x^i (i!(i+1)!)^{-1/2}$

Question

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?

Answer 1

Like the comments mentioned, the sequence can be rewritten as

$$S(x) := \sum_{i=0}^{\infty} \frac{x^i}{\sqrt{i!(i+1)!}} = \sum_{i=0}^{\infty} \frac{x^i}{\sqrt{i!i!(i+1)}} = \sum_{i=0}^{\infty} \frac{x^i}{i!\sqrt{i+1}}$$

From here, I was able to give an approximation for this sum, using these lower and upper bounds:

$$\sum_{i=0}^{\infty} \frac{x^i}{i!\sqrt{i+1}} \le \sum_{i=0}^{\infty} \frac{x^i}{i!} = e^x$$

And

$$\sum_{i=0}^{\infty} \frac{x^i}{i!\sqrt{i+1}} \ge \sum_{i=0}^{\infty} \frac{x^i}{i!(i+1)} = \sum_{i=0}^{\infty} \frac{x^i}{(i+1)!} = \frac{1}{x}\sum_{i=1}^{\infty} \frac{x^i}{i!} = \frac{1}{x}\left(\sum_{i=0}^{\infty} \frac{x^i}{i!} - 1\right) = \frac{e^x-1}{x}$$

So the sum always goes between the values

$$\frac{e^x-1}{x} \le S(x) \le e^x$$

From this, I guessed that maybe

$$S(x) \stackrel{?}{\approx} \frac{e^x}{\sqrt{x}}$$

And it turns out, yes, asymptotically, this indeed holds! I verified this using Python and MatPlotLib. I plotted $S(x)$ (summed up to $100$) in red and the $\frac{e^x}{\sqrt{x}}$ approximation in blue. The two plots are for the case $x \in [0,3]$ and $x \in [0,6]$. Only the begininning portion, about $x \in [0,0.5]$ is off.

I don't think it's easy to find a closed formula for this, WolframAlpha just gives up for example. So this might be the best we can get.

Of course, we might even be able to get better low-end approximations too with functions like

$$\frac{e^x - 1}{\sqrt{x}}, \quad \frac{e^x - x}{\sqrt{x}}, \quad \frac{e^x}{\sqrt{x}} - x$$

Which I have tested, but none were significantly better than $\frac{e^x}{\sqrt{x}}$.

EDIT: Upon further trial and error, $S(x) \approx \frac{e^x}{\sqrt{x+1}}$ might be even better, this one's even closer on the low end:

EDIT $2$: I found this too. This is $S(x) \approx \frac{e^x + \sqrt{x}}{\sqrt{x+1}}$. This fits better on the high end, but worse on the low end.

Answer 2

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: