(Not to be confused with Catalan's constant)
Evaluate the infinite sum $$S = \sum_{n=0}^\infty \frac{(-1)^{T_n}}{(2n+1)^2} = 1 – \frac1{3^2} – \frac1{5^2} + \frac1{7^2} + \frac1{9^2} – \frac1{11^2} – \frac1{13^2} + \frac1{15^2} + \cdots$$
where $T_n = \frac{n(n+1)}2$ is the $n$-th triangular number.
Equivalently, with $\omega = e^{i\frac\pi4}$, we can write the series as
$$S = \frac1{2\sqrt2} \sum_{n=1}^\infty \frac{\omega^n – \omega^{3n} – \omega^{5n} + \omega^{7n}}{n^2}$$
or
$$S = \sum_{n=1}^\infty \frac{a_n}{n^2} \quad \text{ where } \quad \begin{cases}a_{8n}=a_{8n+2}=a_{8n+4}=a_{8n+6}=0\\a_{8n+1}=a_{8n+7}=+1\\a_{8n+3}=a_{8n+5}=-1\end{cases}$$
I see a linear combination of $\mathrm{Li}_2(\cdot)$ terms but I'm not aware of or otherwise familiar with any identities that might be helpful in simplifying the sum.
Borrowing inspiration from some other questions, I am hoping to cook up a periodic function $f(x)$ so that I can exploit its Fourier series to evaluate $S$. Or possibly the sums that make up $S$.
For instance, I've started the hunt with the cosine expansion of $f(x)=\cos\left(\frac x8\right)$ on $[-\pi,\pi]$, given by
$$\cos\left(\frac x8\right) = \frac8\pi \sin\left(\frac\pi8\right) – \frac8\pi \sin\left(\frac\pi8\right) \sum_{n=1}^\infty \left(\frac{(-1)^n}{8n-1} – \frac{(-1)^n}{8n+1}\right) \cos(nx)$$
as well as $\cos\left(\frac{3x}8\right)$, which has an expansion containing denominators with $8n\pm3$. I don't know whether having $8n-1$ and $8n-3$ will be a problem, since there is some correspondence between $8n-1$ and $8n+7$, as well as $8n-3$ and $8n+5$.
For posterity, I also considered the sine expansion of $f$ as well as the co/sine expansions of $\sin\left(\frac x8\right)$.
$$\begin{align*}
\cos\left(\frac x8\right) &= \frac8\pi \sum_{n=1}^\infty \left(\frac{1-(-1)^n \cos\left(\frac\pi8\right)}{8n-1} + \frac{1-(-1)^n \cos\left(\frac\pi8\right)}{8n+1}\right) \sin(nx) \\[1ex]
\sin\left(\frac x8\right) &= \frac{16}\pi \sin^2\left(\frac\pi{16}\right) – \frac8\pi \sum_{n=1}^\infty \left(\frac{1-(-1)^n \cos\left(\frac\pi8\right)}{8n-1} – \frac{1 – (-1)^n \cos\left(\frac\pi8\right)}{8n+1}\right) \cos(nx)\\[1ex]
\sin\left(\frac x8\right)&= -\frac8\pi \sin\left(\frac\pi8\right) \sum_{n=1}^\infty \left(\frac{(-1)^n}{8n-1} + \frac{(-1)^n}{8n+1}\right) \sin(nx)
\end{align*}$$
Next I looked at $x \cos\left(\frac x8\right)$ in an attempt to get the squared denominators, with e.g. cosine expansion
$$\begin{align*}
x\cos\left(\frac x8\right) &= \frac8\pi \left(8\cos\left(\frac\pi8\right)+\pi\sin\left(\frac\pi8\right)-8\right) \\ & \qquad – 8\sin\left(\frac\pi8\right) \sum_{n=1}^\infty \left(\frac{(-1)^n}{8n-1} – \frac{(-1)^n}{8n+1}\right) \cos(nx) \\ & \qquad – \frac{64}\pi \sum_{k=1}^n \left(\frac{1 – (-1)^n \cos\left(\frac\pi8\right)}{(8n-1)^2} + \frac{1 – (-1)^n \cos\left(\frac\pi8\right)}{(8n+1)^2}\right) \cos(nx) \\[1ex]
(x-\pi) \cos\left(\frac x8\right) &= \frac{64}\pi \left(\cos\left(\frac\pi8\right)-1\right) – \frac{64}\pi \sum_{n=1}^\infty \left(\frac{1 – (-1)^n \cos\left(\frac\pi8\right)}{(8n-1)^2} + \frac{1 – (-1)^n \cos\left(\frac\pi8\right)}{(8n+1)^2}\right) \cos(nx)
\end{align*}$$
The next move would be to pick a value of $x$ to recover some numerical series, but the right choice – if there is one – isn't obvious to me. I can nearly recover two that I want, namely
$$\sum_{n=0}^\infty \left(\frac1{(8n+1)^2} + \frac1{(8n+7)^2}\right)$$
but still have to deal with their alternating variants. Presumably I can do something similar with $\cos\left(\frac{3x}8\right)$, but one thing at a time.
I'm open to other methods for evaluating the series, but I'm more interested in the Fourier approach, if it's tenable.
Best Answer
For reference, here is a solution using dilogarithm identities $$\newcommand{\dilog}{\operatorname{Li}_2}\dilog(z)+\dilog(-z)=\frac12\dilog(z^2),\qquad\dilog(z)+\dilog(z^{-1})=-\frac{\pi^2}{6}-\frac{\log^2(-z)}{2},$$ with the principal value of the logarithm.
Starting with the identity from the OP (with $\omega=e^{i\pi/4}$): \begin{align*} 2S\sqrt2&=\dilog(\omega)-\dilog(\omega^3)-\dilog(\omega^5)+\dilog(\omega^7) \\&=\dilog(\omega)-\dilog(-\omega)+\dilog(\omega^{-1})-\dilog(-\omega^{-1}) \\&=2\dilog(\omega)-\frac12\dilog(i)+2\dilog(\omega^{-1})-\frac12\dilog(-i) \\&=-\frac32\frac{\pi^2}{6}-\underbrace{\log^2(-\omega)}_{=(-3i\pi/4)^2}+\frac14\underbrace{\log^2(-i)}_{=(-i\pi/2)^2}=\frac{\pi^2}{4}. \end{align*} Thus $S=\pi^2/(8\sqrt{2})$.