What is the value of $\frac11+\frac13-\frac15-\frac17+\frac19+\frac1{11}-\dots$? The series $\sum_{k=1}^{\infty }\frac{(-1)^{k+1}}{2k-1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots$ converges to $\frac{\pi}{4}$. Here, the sign alternates every term.
The series $\displaystyle\sum_{k=1}^{\infty }{(-1)^{\left(k^{2} + k + 2\right)/2} \over 2k-1}=\frac{1}{1}+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\dots$ also converges. Here, the sign alternates every two terms.

What is the convergence value, explicitly, of the second series?

The first summation is noted above, because it might be a useful information to evaluate the second summation.

Hint:
$$\frac{1}{1}+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\dots=\sum_{k=1}^{\infty }\frac{1}{8k-7}+\frac{1}{8k-5}-\frac{1}{8k-3}-\frac{1}{8k-1}$$
$$=\sum_{k=1}^{\infty }(\frac{1}{8k-7}-\frac{1}{8k-1})+(\frac{1}{8k-5}-\frac{1}{8k-3})$$
$$=\sum_{k=1}^{\infty }(1+\frac{1}{8k+1}-\frac{1}{8k-1})+(\frac{1}{3}+\frac{1}{8k+3}-\frac{1}{8k-3})$$
$$=\frac{4}{3}+\sum_{k=1}^{\infty }(\frac{-2}{64k^2-1}+\frac{-6}{64k^2-9})$$
$$=\frac{4}{3}-\frac{1}{32}\sum_{k=1}^{\infty }\frac{1}{k^2-\frac{1}{64}}-\frac{3}{32}\sum_{k=1}^{\infty }\frac{1}{k^2-\frac{9}{64}}$$
then use
$$\frac{1-\pi x \cot(\pi x)}{2x^2}=\sum_{k=1}^{\infty }\frac{1}{k^2-x^2}$$
