Infinite series involving $\sqrt{n}$

Solution 1:

As indicated by Aryabhata the first result is simply a polylogarithm : $$\operatorname{Li}_s(z)=\sum_{k=1}^\infty \frac {z^k}{k^s}$$ for the specific value $s=-\frac 12$

Note that the Lerch zeta function allows to obtain $\sqrt{n+\alpha}$ coefficients.

For the second one let's search an approximation of $\ \displaystyle S(z):=\sum_{n=1}^{\infty}\frac{\sqrt{n}\,z^n}{n!}$ as $z\to +\infty$

The asymptotic expansion indicated in the comment from this SE thread (setting $z:=\frac te$) was : $$\frac 1{\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{(ez)^n}{n^n} \sim \sum_{n=1}^{\infty}\frac{\sqrt{n}\,z^n}{n!}\sim \sqrt{z}\ e^z\quad\text{as}\ z\to +\infty$$

The term at the left may be obtained with the Stirling approximation $\;\displaystyle n!\sim \sqrt{2\pi n}\left(\frac ne\right)^n$.

The term at the right was of the 'wild guess kind' : since $\displaystyle\frac{n!}{\sqrt{n}}$ is of order $\left(n-\frac 12\right)!\ $ (i.e. $\Gamma\left(n+\frac 12\right)$) for $n\gg1$ and since the small values of $n$ will contribute very little for $z\gg 1$ we will have $\ \displaystyle S(z)\approx\sqrt{z}\sum_{n=1}^{\infty}\frac{z^{n-\frac 12}}{\left(n-\frac 12\right)!}\approx \sqrt{z}\;e^z$

Let's try to put this on more solid ground. Using the Stirling series we get : \begin{align} \sqrt{n}\;\Gamma\left(n+\frac 12\right)&= n!\left(1-\frac 1{8\;n}+\frac 1{128\;n^2}+\frac 5{1024\;n^3}+O\left(\frac 1{n^4}\right) \right)\\ &=n!\left(1-\frac 1{8\left(n+\frac 12\right)}-\frac 7{128\left(n+\frac 12\right)\left(n+\frac 32\right)}-\frac {75}{1024\left(n\cdots+\frac 52\right)}+O\left(\frac 1{n^4}\right) \right)\\ \end{align} and : $$\sum_{n=1}^{\infty}\frac{\sqrt{n}\,z^n}{n!}= \sqrt{z}\sum_{n=1}^{\infty}\frac {z^{n-\frac 12}}{\Gamma\left(n+\frac 12\right)} \left(1-\frac 1{8\left(n+\frac 12\right)}-\frac 7{128\left(n+\frac 12\right)\left(n+\frac 32\right)}-\cdots \right)$$

Now from a general formula for the incomplete gamma function $\ \displaystyle\gamma(a,z)=z^ae^{-z}\sum_{k=0}^\infty\frac {z^k}{(a)_{k+1}}$ and the specific formula $\;\gamma\bigl(\frac 12,z\bigr)=\sqrt{\pi}\,\operatorname{erf}\bigl(\sqrt{z}\bigr)\,$ ( $\operatorname{erf}$ is the error function) we get : $$\sum_{n=1}^{\infty}\frac {z^{n-\frac 12}}{\Gamma\left(n+\frac 12\right)}=e^z\operatorname{erf}\left(\sqrt{z}\right)$$ In practice for $z\to +\infty$ we get $\ \displaystyle\sum_{n=1}^{\infty}\frac {z^{n-\frac 12}}{\Gamma\left(n+\frac 12\right)}\sim e^z-\frac 1{\sqrt{\pi z}}$

with the result (hiding this last small error in the larger error term) : $$\sum_{n=1}^{\infty}\frac{\sqrt{n}\,z^n}{n!}= \sqrt{z}\ e^z\left[1-\frac 1{8\,z}-\frac 7{128\,z^2}-\frac {75}{1024\,z^3}+O\left(\frac 1{z^4}\right)\right]$$

Concerning the general expression perhaps the fractional derivative $D^{1/2}$ may help...
(Update: the idea was that $D(x^n)=n\;x^{n-1}$ so that $D^{1/2}$ could be $\sqrt{n}\,x^{\cdots}$ but I fear this doesn't work since the power of $x$ is changed after the first 'half-derivative'...)