How prove this sum$\sum_{k=0}^{\infty}\frac{(-1)^{\frac{k(k+1)}{2}}}{(2k+1)^2}=\frac{\sqrt{2}\pi^2}{16}$

Question

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

Answer 1

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

$$\frac{\pi^2}{128} 2 \left (\frac{2+\sqrt{2}}{1/2} - \frac{2-\sqrt{2}}{1/2} \right ) = \frac{\sqrt{2} \pi^2}{16}$$

Answer 2

Hint: Split the sum into four parts $$ \sum_{k=0}^\infty\frac{(-1)^{\frac{k(k+1)}{2}}}{(2k+1)^2}= \sum_{k=0}^\infty\frac{1}{(8k+1)^2} -\sum_{k=0}^\infty\frac{1}{(8k+3)^2} -\sum_{k=0}^\infty\frac{1}{(8k+5)^2} +\sum_{k=0}^\infty\frac{1}{(8k+7)^2} $$

Answer 3

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 $\dfrac{\sqrt{2}\pi^2}{16}$