How prove this sum$\sum_{k=0}^{\infty}\frac{(-1)^{\frac{k(k+1)}{2}}}{(2k+1)^2}=\frac{\sqrt{2}\pi^2}{16}$ How show that
$$\sum_{k=0}^{\infty}\dfrac{(-1)^{\frac{k(k+1)}{2}}}{(2k+1)^2}=\dfrac{\sqrt{2}\pi^2}{16}$$
My idea:I know how to  prove  the following sum
$$\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)^2}=\int_{0}^{1}\dfrac{\arctan{x}}{x}dx=C$$
where the $C$ is Catalan constant
But for this$$\sum_{k=0}^{\infty}\dfrac{(-1)^{\frac{k(k+1)}{2}}}{(2k+1)^2}=\dfrac{\sqrt{2}\pi^2}{16}$$I can't,Thank you for  you help
 A: Hint: Split the sum into four parts
$$
\sum_{k=0}^\infty\frac{(-1)^{\frac{k(k+1)}{2}}}{(2k+1)^2}=
\sum_{k=0}^\infty\frac{1}{(8k+1)^2}
-\sum_{k=0}^\infty\frac{1}{(8k+3)^2}
-\sum_{k=0}^\infty\frac{1}{(8k+5)^2}
+\sum_{k=0}^\infty\frac{1}{(8k+7)^2}
$$
A: The pattern of the signs of the sum are
$$a_0-a_1-a_2+a_3+a_4-a_5-a_6+a_7+a_8-\cdots$$
so that each term of the form $8 k+3$ and $8 k+5$ is negative, and $8 k+1$ and $8 k+7$ is positive.  Thus we may write the sum as
$$\sum_{k=0}^{\infty} \left [\frac1{(8 k+1)^2} - \frac1{(8 k+3)^2} - \frac1{(8 k+5)^2}+\frac1{(8 k+7)^2} \right ]$$
Note that this sum is symmetric in sign about $k$; thus we may write the sum as 
$$\frac12 \sum_{k=-\infty}^{\infty} \left [\frac1{(8 k+1)^2} - \frac1{(8 k+3)^2} - \frac1{(8 k+5)^2}+\frac1{(8 k+7)^2} \right ]$$
We may thus evaluate this sum using the residue theorem.  Omitting details, we use the result 
$$\sum_{k=-\infty}^{\infty} f(k) = -\sum_n \operatorname*{Res}_{z=z_n} \pi \, \cot{(\pi z)} \, f(z)$$
where
$$f(z) = \frac12 \left [\frac1{(8 z+1)^2} - \frac1{(8 z+3)^2} - \frac1{(8 z+5)^2}+\frac1{(8 z+7)^2} \right ]$$
$f$ has (double) poles at $z_1=-1/8$, $z_2=-3/8$, $z_3=-5/8$, and $z_4=-7/8$.  Thus, for instance,
$$\begin{align}\operatorname*{Res}_{z=-1/8} \pi \, \cot{(\pi z)} \, f(z) &= \frac12\pi \frac1{8^2}\left [\frac{d}{dz} \left ( \left [1- \frac{(8 z+1)^2}{(8 z+3)^2} - \frac{(8 z+1)^2}{(8 z+5)^2}+\frac{(8 z+1)^2}{(8 z+7)^2} \right ] \cot{(\pi z)} \right ) \right ]_{z=-1/8} \\ &= -\frac1{128} \pi^2 \csc^2{\frac{\pi}{8}}\end{align}$$
The factor of $1/8^2$ comes from the fact that the residue calculation has us multiply $f$ by $(z+1/8)^2$.  Similarly,
$$\operatorname*{Res}_{z=-3/8} \pi \, \cot{(\pi z)} \, f(z) = \frac1{128} \pi^2 \csc^2{\frac{3 \pi}{8}}$$
$$\operatorname*{Res}_{z=-5/8} \pi \, \cot{(\pi z)} \, f(z) = \frac1{128} \pi^2 \csc^2{\frac{5 \pi}{8}}$$
$$\operatorname*{Res}_{z=-7/8} \pi \, \cot{(\pi z)} \, f(z) = -\frac1{128} \pi^2 \csc^2{\frac{7 \pi}{8}}$$
Using
$$\sin^2{\frac{\pi}{8}} = \sin^2{\frac{7\pi}{8}} = \frac{2-\sqrt{2}}{4} = \frac{1/2}{2+\sqrt{2}}$$
$$\sin^2{\frac{3\pi}{8}} = \sin^2{\frac{5\pi}{8}} = \frac{2+\sqrt{2}}{4} = \frac{1/2}{2-\sqrt{2}}$$
Then the sum is finally
$$\frac{\pi^2}{128} 2 \left (\frac{2+\sqrt{2}}{1/2} - \frac{2-\sqrt{2}}{1/2} \right ) = \frac{\sqrt{2} \pi^2}{16}$$
A: I do not have any idea about the help of this but it could, may be, give you ideas. Before replacement the sum is given by $$\frac{1}{64} \left(\psi ^{(1)}\left(\frac{1}{8}\right)-\psi
   ^{(1)}\left(\frac{3}{8}\right)-\psi ^{(1)}\left(\frac{5}{8}\right)+\psi
   ^{(1)}\left(\frac{7}{8}\right)\right)$$ where appear polygamma functions. After replacement, I get your term $\dfrac{\sqrt{2}\pi^2}{16}$
