For what values of $a>0$ is the equation $\sum\limits_{n=1}^{\infty} \frac{2}{(n-0.5)^2+a} = \frac{\pi}{\sqrt{a}}$ true?

Question

For what values of $a>0$ is the equation $$\sum\limits_{n=1}^{\infty} \frac{2}{(n-0.5)^2+a} = \frac{\pi}{\sqrt{a}}$$ true?

I saw this claim here, and know it's not true for all $a$ since the LHS converges while the RHS diverges as $a \to 0.$ However, as $a$ increases we seem to get a better and better approximation.

This answer provides the helpful formula $$\sum_{n = -\infty}^{\infty} \frac{1}{(n+\alpha)^2+\gamma^2} = \frac{\pi}{\gamma} \frac{\sinh{2\gamma \pi}}{\cosh{2\gamma \pi}-\cos{2\alpha \pi}}.$$ Unfortunately, $a$ may not be an integer, so we can't just plug in $\alpha = -0.5, \gamma = a$ and hope to fold the sums over negative and positive integers over each other. Any other ideas?

Answer 1

We start from the result linked in the question: $$ \sum_{n = -\infty}^{\infty} \frac{1}{(n+\alpha)^2+\gamma^2} = \frac{\pi}{\gamma} \frac{\sinh{2\gamma \pi}}{\cosh{2\gamma \pi}-\cos{2\alpha \pi}}. $$ For the specific case of $\alpha = -\frac{1}{2}$, it is not too hard to see that $$ \sum_{n = -\infty}^{0} \frac{1}{(n-\frac12)^2+\gamma^2} = \sum_{n = 1}^{\infty} \frac{1}{(n-\frac12)^2+\gamma^2}, $$ and so we have $$ 2 \sum_{n = 1}^{\infty} \frac{1}{(n-\frac12)^2+\gamma^2} = \sum_{n = -\infty}^{\infty} \frac{1}{(n-\frac12)^2+\gamma^2} \\=\frac{\pi}{\gamma} \frac{\sinh{2\gamma \pi}}{\cosh{2\gamma \pi}-\cos{ \pi}} = \frac{\pi}{\gamma} \frac{\sinh{2\gamma \pi}}{\cosh{2\gamma \pi}+ 1}. $$ But $\sinh 2x = 2 \sinh x \cosh x$ and $\cosh 2x + 1 = 2 \cosh^2 x$, so $$ \sum_{n = 1}^{\infty} \frac{2}{(n-\frac12)^2+\gamma^2} = \frac{\pi}{\gamma} \tanh{\gamma \pi} = \frac{\pi}{\sqrt{a}} \tanh\left(\sqrt{a} \pi\right), $$ where we have identified $\gamma = \sqrt{a}$ in the last step. As the OP notes, this should have a finite limit as $a \to 0$, and a bit of work shows that this limit is indeed finite and equal to $\pi^2$.

Thus, the given formula is true for all values of $a$ for which $\tanh(\sqrt{a} \pi) = 1$, which is... none of them. For all real $x$, $-1< \tanh(x) < 1$. However, the hyperbolic tangent function does asymptotically approach 1; in fact, asymptotically we have $$ \tanh(\sqrt{a} \pi) \approx 1 - 2 e^{- 2\sqrt{a} \pi}. $$ Even for $a = 1$, we have $\tanh(\pi) \approx 0.996...$ and so the result holds to within less than 1%. It is not too hard to envision an amateur numerologist stumbling upon this relation "experimentally" with a computer and thinking that it must be true. (I note that the given equation was posted on viXra, which does have a reputation for such things.) It is a remarkably good approximation, particularly for large values of $a$, but it is not in fact true.

Answer 2

Consider the partial sum $$S_p=\sum_{n=1}^p \frac{2}{\left(n-\frac{1}{2}\right)^2+a}=\sum_{n=1}^p \frac{2}{ \left(n+\frac{1}{2} \left(-2 \sqrt{-a}-1\right)\right) \left(n+\frac{1}{2} \left(2 \sqrt{-a}-1\right)\right)}$$ Use partial fraction decomposition and get $$S_p=\frac{\psi \left(p-\sqrt{-a}+\frac{1}{2}\right)-\psi \left(p+\sqrt{-a}+\frac{1}{2}\right)-\psi \left(\frac{1}{2}-\sqrt{-a}\right)+\psi \left(\sqrt{-a}+\frac{1}{2}\right)}{\sqrt{-a}}$$ Now, using the asymptotics $$\psi(n)=\log (n)-\frac{1}{2 n}-\frac{1}{12 n^2}+O\left(\frac{1}{n^4}\right)$$ Using it twice and continuing with Taylor series $$S_p=\frac{\psi \left(\frac{1}{2}+\sqrt{-a}\right)-\psi \left(\frac{1}{2}-\sqrt{-a}\right)}{\sqrt{-a}}-\frac 2 p+\frac{4 a+1}{6 p^3}+O\left(\frac{1}{p^5}\right)$$ and using now $$\psi \left(\frac{1}{2}+b\right)-\psi \left(\frac{1}{2}-b\right)=\pi \tan (\pi b)$$ thus $$\frac{\psi \left(\frac{1}{2}+\sqrt{-a}\right)-\psi \left(\frac{1}{2}-\sqrt{-a}\right)}{\sqrt{-a}}=\frac{\pi }{\sqrt{a}} \tanh \left(\pi \sqrt{a}\right)$$ This is what Wolfram Alpha provided.