The sum we want is
[tex]\displaystyle \sum_{n=0}^\infty \frac{(-1)^{T_n}}{(2n+1)^2} = 1 - \frac1{3^2} - \frac1{5^2} + \frac1{7^2} + \cdots[/tex]
where [tex]T_n=\frac{n(n+1)}2[/tex] is the n-th triangular number, with a repeating sign pattern (+, -, -, +). We can rewrite this sum as
[tex]\displaystyle \sum_{k=0}^\infty \left(\frac1{(8k+1)^2} - \frac1{(8k+3)^2} - \frac1{(8k+7)^2} + \frac1{(8k+7)^2}\right)[/tex]
For convenience, I'll use the abbreviations
[tex]S_m = \displaystyle \sum_{k=0}^\infty \frac1{(8k+m)^2}[/tex]
[tex]{S_m}' = \displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{(8k+m)^2}[/tex]
for m ∈ {1, 2, 3, …, 7}, as well as the well-known series
[tex]\displaystyle \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}[/tex]
We want to find [tex]S_1-S_3-S_5+S_7[/tex].
Consider the periodic function [tex]f(x) = \left(x-\frac12\right)^2[/tex] on the interval [0, 1], which has the Fourier expansion
[tex]f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi nx)}{n^2}[/tex]
That is, since f(x) is even,
[tex]f(x) = a_0 + \displaystyle \sum_{n=1}^\infty a_n \cos(2\pi nx)[/tex]
where
[tex]a_0 = \displaystyle \int_0^1 f(x) \, dx = \frac1{12}[/tex]
[tex]a_n = \displaystyle 2 \int_0^1 f(x) \cos(2\pi nx) \, dx = \frac1{n^2\pi^2}[/tex]
(See attached for a plot of f(x) along with its Fourier expansion up to order n = 10.)
Expand the Fourier series to get sums resembling the [tex]S'[/tex]-s :
[tex]\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{k=0}^\infty \frac{\cos(2\pi(8k+1) x)}{(8k+1)^2} + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+2) x)}{(8k+2)^2} + \cdots \right. \\ \,\,\,\, \left. + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+7) x)}{(8k+7)^2} + \sum_{k=1}^\infty \frac{\cos(2\pi(8k) x)}{(8k)^2}\right)[/tex]
which reduces to the identity
[tex]\pi^2\left(\left(x-\dfrac12\right)^2-\dfrac{21}{256}\right) = \\\\ \cos(2\pi x) {S_1}' + \cos(4\pi x) {S_2}' + \cos(6\pi x) {S_3}' + \cos(8\pi x) {S_4}' \\\\ \,\,\,\, + \cos(10\pi x) {S_5}' + \cos(12\pi x) {S_6}' + \cos(14\pi x) {S_7}'[/tex]
Evaluating both sides at x for x ∈ {1/8, 3/8, 5/8, 7/8} and solving the system of equations yields the dependent solution
[tex]\begin{cases}{S_4}' = \dfrac{\pi^2}{256} \\\\ {S_1}' - {S_3}' - {S_5}' + {S_7}' = \dfrac{\pi^2}{8\sqrt 2}\end{cases}[/tex]
It turns out that
[tex]{S_1}' - {S_3}' - {S_5}' + {S_7}' = S_1 - S_3 - S_5 + S_7[/tex]
so we're done, and the sum's value is [tex]\boxed{\dfrac{\pi^2}{8\sqrt2}}[/tex].
