# Infinite series involving $\sqrt{n}$

I am looking for examples of infinite series, whose sum is expressed as distributions or known functions, with a $\sqrt{n}$ in each term, such as:

$$\sum_{n=0}^{\infty} \sqrt{n} z^n, \quad \sum_{n=0}^{\infty} \frac{\sqrt{n}}{n!} z^n,$$ or more generally:

$$\sum_{n=0}^{\infty} \sqrt{n} f(n) z^n$$ where $f(n)$ is a combination (product, division) of integer factorials, constants, etc...

Here, $z$ can be real or complex.

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Have you tried wolfram alpha ? –  Aryabhata May 1 '13 at 8:56
Well, I just tried, but it is difficult fo formulate the question such as wolfram alpha understands it.... –  Trimok May 1 '13 at 9:14
Well, you have to make guesses for $f(n)$. Don't expect to get results for every try. At least we get one for $f(n)=1$. –  Aryabhata May 1 '13 at 9:21
Oh yes, Ok, thanks, the result is $Li_{-1/2}(z) for |z| < 1$ –  Trimok May 1 '13 at 9:24
For the second one there is a clear relation with $\displaystyle \sum_{k=1}^{\infty}\frac{t^{k}}{k^{k}}$ as discussed in this SE thread. –  Raymond Manzoni May 1 '13 at 9:28

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'...)

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Great. Thanks . –  Trimok May 2 '13 at 8:41
Thanks @Trimok ! Let's just add that the polylogarithm allows to handle the general problem of $\ \displaystyle\sum_{n=0}^{\infty} \sqrt{n}\;P(n)\;z^n\$ with $P(n)$ a polynomial since $\ \displaystyle\sum_{n=0}^{\infty} \sqrt{n}\;n^k\;z^n=\operatorname{Li}_{-(k+1/2)}(z)$. –  Raymond Manzoni May 2 '13 at 9:46
Yes, I have noticed that. I wonder if there is a general theorem, that when z goes to infinity, then : $\sum_{n=1}^{\infty}\frac{f(n)\,z^n}{n!} \sim f(z) e^z$, for some smooth and increasing function f –  Trimok May 3 '13 at 6:52
@Trimok: From my result you may deduce that for a polynomial $P(z)$ (not $P(n)$ sorry) we have : $\ \displaystyle \sum_{n=1}^{\infty}\frac{\sqrt{n}\;P(z)\,z^n}{n!}\sim \sqrt{z}\ P(z)\ e^z\left[1-\frac 1{8\,z}-\frac 7{128\,z^2}-\frac {75}{1024\,z^3}+O\left(\frac 1{z^4}\right)\right]$ (note that these coefficients are as important as the decreasing coefficients of the polynomial). This may probably be expanded further to functions $f(n)$ expandable as Laurent series (I think that the fractional power $n^{\alpha}$ will introduce coefficients depending of $\alpha$ that is $\frac 12$ here). –  Raymond Manzoni May 3 '13 at 12:31
@Trimok: If you want to do this for $P(n)$ you may use $\ \displaystyle \sum_{n=1}^{\infty}\frac{\sqrt{n}\;n^k\,z^n}{n!}=\theta^k \sum_{n=1}^{\infty}\frac{\sqrt{n}\,z^n}{n!}$ with $\theta=z\frac d{dz}$ the 'theta operator' and you should get something like $\ \displaystyle \sum_{n=1}^{\infty}\frac{\sqrt{n}\;P(n)\,z^n}{n!}\sim P\left(z\frac d{dz}\right)\left[\sqrt{z}\ e^z\left[1-\frac 1{8\,z}-\frac 7{128\,z^2}-\frac {75}{1024\,z^3}+O\left(\frac 1{z^4}\right)\right]\right]$ (more terms may be necessary in this case!). –  Raymond Manzoni May 3 '13 at 12:41