A "relative" of the exponential function

It is well known that $$\mathrm{e}=\sum_{n=0}^\infty \frac{1}{n!}.$$ Recently, I was reminded that the volume of an $n$-simplex in $n$-dimensional space, with vertices $v_0,v_1,\dots,v_n$ is $$\left| \frac{1}{n!} \det \left(v_1-v_0,v_2-v_0,\dots,v_n-v_0 \right) \right|.$$ In particular, if for all $i$, $v_i-v_0=e_i$ is the $i$th standard basis vector, we find that the $n$-dimensional volume of the so-called "corner of the unit cube", $\Delta_c^n$ is $1/n!$.

Thus, one can say that Euler's number is the sum of the volumes of corners of unit cubes, taken over all dimensions, or as a formula $$\mathrm{e}=\sum_{n=0}^\infty \operatorname{vol_n}\left( \Delta_c^n \right).$$

Moreover, scaling each of the $\Delta_c^ns$ by the same nonnegative real number $x$, scales its volume by $x^n$, and one gets a geometrical formula for the exponential function:

$$\mathrm{e}^x =\sum_{n=0}^\infty \operatorname{vol_n} \left( x \cdot \Delta_c^n \right)=\sum_{n=0}^\infty \operatorname{vol_n} \left( \Delta_c^n \right) x^n. $$ In fact, the power series on the right can be continued analytically to a complex, entire function of $x$.

One can do the same for other families of shapes, for example it is known that the volume of an $n$-dimensional ball, of radius $x$, is $x^n \pi^{n/2}/\Gamma(n/2+1)$. Hence, the associated function takes the form $$\sum_{n=0}^\infty \frac{\pi^\frac{n}{2}}{\Gamma\left(\frac{n}{2}+1 \right)} x^n =e^{\pi x^2} \left(\text{erf}\left(\sqrt{\pi } x\right)+1\right).$$

I was wondering what happens if one replaces the corners of the cubes of the exponential function by the regular simplices, of side-length $x$. The associated function is then $$f(x)=\sum_{n=0}^\infty \frac{\sqrt{1+n}}{n! 2^\frac{n}{2}} x^n,$$ which is entire. I've tried expressing $f(x)$ in terms of other functions, to no avail. Moreover, I couldn't even get a closed form for the simple-looking sum $$f\left( \sqrt{2} \right)= \sum_{n=0}^\infty \frac{\sqrt{1+n}}{n!}.$$ I would appreciate any help understanding/simplifying $f(x)$ in general, and $f(\sqrt{2})$ in particular. Thank you!

Solution 1:

$$ \sum_{n=0}^{\infty} \frac{\sqrt{1+n}}{n!}=\sum_{n=0}^{\infty} \frac{\sqrt{n(1+1/n)}}{n!}$$ $$ \sum_{n=0}^{\infty} \frac{\sqrt{n(1+1/n)}}{n!}=\sum_{n=0}^{\infty} \frac{\sqrt{n}\sqrt{1+1/n}}{n!}$$ and then that square root can be expanded hypergeometrically.

Maybe you want 3j/6j/9j/12j symbols?