Value of $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n!}}$? I was reading this old question and fascinated by the second infinite sum $$\sum_{n=1}^{\infty}\dfrac{1}{\sqrt{n!}}.$$ This is clearly convergent (by comparison or ratio test) and, we can obtain some crude approximations of this using inequalities like $n^4\le n!\le (n!)^2,$ here the first inequality holds for all $n\ge7.$ But, I wonder if we can find the exact value of this series using (familiar) special functions/constants. How would you attack to this series?
COMMENT:

It can be shown that:

$$∑^∞_{n=1} \frac{1}{n!}&lt;3$$:

$$1+1+\frac{1}{2}+\frac{1}{2\times 3}+\frac{1}{2\times 3\times 4}+ . . . \frac{1}{2\times 3\times . . .\times n}&lt;1+\big(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ . . . +\frac{1}{2^{n-1}}\big)=1+\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}=1+2(1-\frac{1}{2^n})=3-\frac{1}{2^{n-1}}&lt;3$$

We may conclude that:

$$\sqrt{∑^∞_{n=1} \frac{1}{n!}}&lt;\sqrt 3=1.73...$$

We can see that:

$$\sqrt{∑^∞_{n=1} \frac{1}{n!}}&lt;∑^∞_{n=1} \frac{1}{\sqrt{n!}}$$

$$\sqrt 3=1.73 &lt;∑^∞_{n=1} \frac{1}{\sqrt{n!}}  &lt;3$$

I checked it up to n=8 which gives $2.46$ , seems to be reasonable.
COMMENT:

It can be shown that:

$$∑^∞_{n=1} \frac{1}{n!}&lt;3$$:

$$1+1+\frac{1}{2}+\frac{1}{2\times 3}+\frac{1}{2\times 3\times 4}+ . . . \frac{1}{2\times 3\times . . .\times n}&lt;1+\big(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ . . . +\frac{1}{2^{n-1}}\big)=1+\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}=1+2(1-\frac{1}{2^n})=3-\frac{1}{2^{n-1}}&lt;3$$

We may conclude that:

$$\sqrt{∑^∞_{n=1} \frac{1}{n!}}&lt;\sqrt 3=1.73...$$

We can see that:

$$\sqrt{∑^∞_{n=1} \frac{1}{n!}}&lt;∑^∞_{n=1} \frac{1}{\sqrt{n!}}$$

$$\sqrt 3=1.73 &lt;∑^∞_{n=1} \frac{1}{\sqrt{n!}}  &lt;3$$

I checked it up to n=8 which gives $2.46$ , seems to be reasonable.