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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.

Elmar Zander
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Trimok
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    Have you tried wolfram alpha ? – Aryabhata May 01 '13 at 08:56
  • Well, I just tried, but it is difficult fo formulate the question such as wolfram alpha understands it.... – Trimok May 01 '13 at 09:14
  • 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 01 '13 at 09:21
  • Oh yes, Ok, thanks, the result is $Li_{-1/2}(z) for |z| < 1$ – Trimok May 01 '13 at 09:24
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    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 01 '13 at 09:28
  • @RaymondManzoni It is very interesting as an approximation, but I don't understand how you factorize the $\sqrt t$ term in $\sqrt{2\pi}\sum_{k=1}^{\infty}\frac{\sqrt{k}(\frac te)^k}{k!}\sim \sqrt{2\pi t}\ e^{\frac te-\frac 12}\ \ \text{as}\ t\to \infty$ – Trimok May 01 '13 at 09:40
  • This is the Lanczos approximation (see the links after that part) and it was initially a guesswork! :-) – Raymond Manzoni May 01 '13 at 09:44
  • (or more exactly I think that the answer is contained in
    Lanczos' paper "A precision approximation for Gamma function" or in Pugh's thesis but didn't verify... I'll do that this afternoon if nobody provides it earlier...)     – Raymond Manzoni May 01 '13 at 09:53

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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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Raymond Manzoni
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  • Great. Thanks . – Trimok May 02 '13 at 08:41
  • 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 02 '13 at 09:46
  • 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 03 '13 at 06:52
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    @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 03 '13 at 12:31
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    @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 03 '13 at 12:41
  • Thanks for your useful remarks – Trimok May 04 '13 at 08:29
  • Glad it helped @Trimok ! – Raymond Manzoni May 04 '13 at 08:39