# How to evaluate infinite series $\sum\limits_{n=0}^\infty\sqrt{B^2+n^2} e^{-an}$

I'm trying to evaluate an infinite series: $$\sum\limits_{n=0}^\infty\sqrt{B^2+n^2} e^{-an}$$ where $a$ and $B$ are real parameters, or equivalently: $$\sum\limits_{n=0}^\infty\sqrt{B^2+n^2} x^n$$ where $0<x<1$.

When $n$ becomes much larger than $B$, the terms will begin to look like: $$nx^n$$ If the entire sum were comprised of these, there's a closed analytic form, which I think is: $$\sum\limits_{n=0}^\infty n x^n = \frac{x}{(1-x)^2}$$

It would be great to know the sum in the first line in closed form though. If anyone could assist either with a closed analytic expression, or even just ideas or techniques as to how to go about evaluating this, It'd be much appreciated.

This series can not be evaluated directly. One approximation method is as follows. Let the Lerch transcendent be defined by $$\phi(z;s,\alpha) = \sum_{n=0}^{\infty} \frac{z^{n}}{(n+\alpha)^{s}}.$$ The series expansion of $\sqrt{1+x}$ is, for the first few terms, $$\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^{2}}{8}+ \frac{3 \, x^{3}}{16} - \cdots.$$ and leads to \begin{align} S &= \sum_{n=0}^{\infty} \sqrt{b^{2}+n^{2}} \, e^{-an} = \sqrt{b^{2}} + \sum_{n=1}^{\infty} n \, \sqrt{1 + \frac{b^{2}}{n^{2}}} \, e^{-an} \\ &= \sqrt{b^{2}} + \sum_{n=1}^{\infty} n \, e^{-an} \, \left( 1 + \frac{b^{2}}{2 \, n^{2}} - \frac{b^{4}}{8 \, n^{4}} + \frac{3 \, b^{6}}{16 \,n^{6}} - \cdots \right) \\ &= \sqrt{b^{2}} - \frac{d}{da} \, \left( \frac{e^{-a}}{1 - e^{-a}} \right) + \frac{b^{2}}{2} \, \sum_{n=1}^{\infty} \frac{e^{-an}}{n} - \frac{b^{4}}{8} \, \sum_{n=0}^{\infty} \frac{e^{-a(n+1)}}{(n+1)^{3}} + \frac{3 \, b^{6}}{16} \, \sum_{n=0}^{\infty} \frac{e^{-a(n+1)}}{(n+1)^{5}} - \cdots \\ &= \sqrt{b^{2}} - \frac{d}{da} \, \left( \frac{1}{e^{a} - 1} \right) - \frac{b^{2}}{2} \, \ln(1-e^{-a}) - \frac{b^{4} \, e^{-a}}{8} \, \phi(e^{-a}; 3,1) + \frac{3 \, b^{6} \, e^{-a}}{16} \, \phi(e^{-a}; 5,1) - \cdots \\ &= \sqrt{b^{2}} + \frac{e^{a}}{(e^{a}-1)^{2}} - \frac{b^{2}}{2} \, \ln(1 - e^{-a}) - \frac{b^{4} \, e^{-a}}{8} \, \phi(e^{-a}; 3,1) + \frac{3 \, b^{6} \, e^{-a}}{16} \, \phi(e^{-a}; 5, 1) - \cdots \end{align}
By using $$\sqrt{1+x} = \sum_{r=0}^{\infty} \frac{(-1)^{r} \, \left(- \frac{1}{2}\right)_{r}}{r!} \, x^{r},$$ where $(x)_{n}$ is the Pochhammer symbol, then \begin{align} S &= \sum_{n=0}^{\infty} \sqrt{b^{2} + n^{2}} \, e^{-an} = \sqrt{b^{2}} + \sum_{n=1}^{\infty} n \, \sqrt{1 + \frac{b^{2}}{n^{2}}} \, e^{-an} \\ &= \sqrt{b^{2}} + \sum_{n=1}^{\infty} n \, e^{-an} + \frac{b^{2}}{2} \, \sum_{n=1}^{\infty} \frac{e^{-an}}{n} + \sum_{r=2}^{\infty} \frac{(-1)^{r} \, \left(- \frac{1}{2}\right)_{r} \, b^{2r}}{r!} \cdot \sum_{n=1}^{\infty} \frac{e^{-an}}{n^{2r-1}} \\ &= \sqrt{b^{2}} - \partial_{a}\left(\frac{1}{e^{a}-1}\right) - \frac{b^{2}}{2} \, \ln(1 - e^{-a}) + \sum_{r=2}^{\infty} \frac{(-1)^{r} \, \left(- \frac{1}{2}\right)_{r} \, b^{2r}}{r!} \, Li_{2r-1}(e^{-a}) \\ &= \sqrt{b^{2}} + \frac{e^{a}}{(e^{a}-1)^{2}} - \frac{b^{2}}{2} \, \ln(1 - e^{-a}) + \sum_{r=2}^{\infty} \frac{(-1)^{r} \, \left(- \frac{1}{2}\right)_{r} \, b^{2r}}{r!} \, Li_{2r-1}(e^{-a}), \end{align} where $Li_{n}(z)$ is the polylogarithm.
• Shouldn't $a$ be evaluated at $0$? – Mark Viola Aug 7 '15 at 16:29
• @Dr.MV The square root was expanded. The derivative with respect to $a$ is one process to evaluate the first sum. – Leucippus Aug 7 '15 at 18:26
• The expansion for the square root should read $1+\frac12 x-\frac1 8x^2+\frac{3}{16}x^3 +\cdots$. Also, all of the $\phi$ terms should have the last argument equal to $0$, not $1$. – Mark Viola Aug 7 '15 at 18:58