Timeline for How to calculate the infinite sum of this double series?
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| Dec 9, 2014 at 14:57 | comment | added | tcya | Hi @HjalmarRosengren, your last step is exactly what I'm trying to prove and idk how. After playing with it I got the double sum in my question. I guess maybe natural numbers are easier to handle so I didn't ask for the original sum of $tanh$. | |
| Dec 9, 2014 at 12:21 | history | edited | Hachino | CC BY-SA 3.0 |
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| Dec 9, 2014 at 9:36 | comment | added | Hachino | Oops, sorry, should've checked twice rather than once. I tried to expand the $\tanh$ using the series of $\frac{1}{1+x}$, which led to the appearance of the $\ln$, but then the result is not obviously equal to $\frac{\log(2)-\pi}4$. | |
| Dec 9, 2014 at 8:39 | comment | added | Hjalmar Rosengren | You made a mistake when separating the sum into even and odd parts. The correct result is $$\sum_{k=0}^\infty\frac 1{(2k+1)^2+m^2}=\frac{\pi\tanh(\pi m/2)}{4m}.$$ This simplifies the double series to the single series $$\sum_{m=1}^\infty\frac{(-1)^m\tanh(\pi m/2)}{m}=\frac{\log(2)-\pi}4. $$ This is convergent (but not absolutely convergent). I am not sure about the simplest way to prove this summation. Probably you can start from the Fourier expansion of an elliptic function and plug in special values, but I am sure there are more elementary ways. | |
| Dec 8, 2014 at 22:01 | comment | added | Alexandre Eremenko | Exactly. The required answer can be obtained only if the sum of this divergent series is given a precise meaning. For this see the Stein and Sharkachi textbook on Complex analysis. | |
| Dec 8, 2014 at 18:22 | comment | added | NAME_IN_CAPS | I don't think that $(-1)^m/\coth(\pi m/2)\rightarrow 0$ as $m\rightarrow\infty$? The denominator tends to 1. | |
| Dec 8, 2014 at 16:45 | history | answered | Hachino | CC BY-SA 3.0 |