I am trying to figure out a practical approach to finding the finite result of an infinite sum - preferably with my HP-50G calculator. For example, if given the following function:
$$i_f(t) = 1.25 + \frac{5}{\pi} \sum_{n=1 (odd)}^{\infty} \left[\frac{\sin 2nt}{n(1 + n^2)} - \frac{\cos 2nt}{1 + n^2} \right]$$
It is of interest to evaluate at $i_f(0)$, so the function becomes:
$$i_f(0) = 1.25 - \frac{5}{\pi} \sum_{n = 1 (odd)}^{\infty} \frac{1}{1 + n^2}$$
My textbook simply states that the summation portion is $0.720$ (i.e. $\sum 1/(1 + n^2) = 0.720$) to three significant figures. A complication is that the summation iterates using consecutive odd integers.
Is there a way to transform this to the point where I can evaluate it with my calculator? WolframAlpha, for instance, shows the following alternate form:
$$\frac{\pi \sinh(\pi)}{4(1 + \cosh(\pi))}$$
How can I simplify it to that alternate form?