MHB Fourier series--showing converges to pi/16

[Dustinsfl]

When a Fourier series contains only sine and cosine terms, evaluating the series isn't too difficult.

However, I want to show a Fourier series with sine and sinh converges to \(\frac{\pi}{16}\).
\[
T(50, 50) = \sum_{n = 1}^{\infty}
\frac{\sin\left(\frac{\pi(2n - 1)}{2}\right)
\sinh\left(\frac{\pi(2n - 1)}{2}\right)}
{(2n - 1)\sinh[(2n - 1)\pi]} = \frac{\pi}{16}
\]
Since this series contains sinh, I am not sure how to evaluate it.

 

[Deveno]

My thought: use

$\text{sinh}(x) = -i \sin(ix)$

 

[Dustinsfl]

My thought: use

$\text{sinh}(x) = -i \sin(ix)$

I think I will still have some trouble though. I end up with:
\[
\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{2n - 1}\frac{\sin\left[\frac{i\pi}{2}(2n-1)\right]}{\sin\left[i\pi(2n-1)\right]}
\]

 

[Opalg]

When a Fourier series contains only sine and cosine terms, evaluating the series isn't too difficult.

However, I want to show a Fourier series with sine and sinh converges to \(\frac{\pi}{16}\).
\[
T(50, 50) = \sum_{n = 1}^{\infty}
\frac{\sin\left(\frac{\pi(2n - 1)}{2}\right)
\sinh\left(\frac{\pi(2n - 1)}{2}\right)}
{(2n - 1)\sinh[(2n - 1)\pi]} = \frac{\pi}{16}
\]
Since this series contains sinh, I am not sure how to evaluate it.

The sin function isn't really there at all, because $\sin\left(\frac{\pi(2n - 1)}{2}\right) = \sin\left(\bigl(n-\frac12\bigr)\pi\right) = (-1)^{n-1}$. Also, you can use the identity $\sinh(2x) = 2\sinh x\cosh x$, to rewrite the sum as $$\sum_{n = 1}^{\infty}\frac{(-1)^n}{2(2n-1)\cosh\left(\bigl(n-\frac12\bigr)\pi\right)}.$$ That looks a bit less complicated than the original series, but I still don't see how to sum it. It clearly converges very fast, because the cosh term in the denominator will get very large after the first few terms. A numerical check shows that the sum of the first three terms is $0.1968518...$, which is very close to $\pi/16$. But that isn't a proof!