For use as an example in a freshman electricity and magnetism class that I'm teaching, I've been trying to come up with an interesting, physically realistic, and mathematically tractable example involving an estimate of the electrical binding energy of a long linear molecule. More details about the physical motivation are here, and it looks like there is no physically realistic system of the type I have in mind. Nevertheless, I am interested now in seeing if I can evaluate the resulting series, which is

$$\sum_{k=1}^\infty \frac{1}{k(k-1/2)}.$$

This smells like the kind of thing that has a closed-form solution, but I haven't been able to figure it out. Any suggestions?

I tried partial fractions, and although that does simplify the form of it, it makes it into the difference of two series that diverge logarithmically. (This actually puts it back into a form that is more directly related to the physical motivation.)

I looked in the big table of series and integrals by Gradshteyn and Ryzhik, who give the similar series

$$ \sum_{k=1}^\infty \frac{1}{k(2k+1)}=2-\ln 2.$$

But they don't seem to say how they got this result (unless there is a footnote that I haven't been able to locate), and I don't see any way to transform my sum into exactly this form by doing things like shifting indices.

For numerical computation, I suspect that convergence would be accelerated by rewriting this as

$$\sum_{k=1}^\infty \left[\frac{1}{k(k-1/2)}-\frac{1}{k^2}\right]+\sum_{k=1}^\infty \frac{1}{k^2},$$

since the sum of $1/k^2$ is easy to look up. But I'm interested in an exact closed-form expression, if possible.

Any ideas?


\begin{align} \sum_{k=1}^{\infty} \frac{1}{k\left(k-\frac{1}{2}\right)}&=2\sum_{k=1}^{\infty} \frac{1}{k(2k-1)}\\ &=4\sum_{k=1}^{\infty}\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\ &=4\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\\ &=4\ln(2). \end{align}

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  • $\begingroup$ D'oh, so obvious now that I see it. Thanks! $\endgroup$ – Ben Crowell Dec 5 '18 at 0:23

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