Using Fourier series to compute $1-\frac1{3^2}-\frac1{5^2}+\frac1{7^2}+\cdots$

Question

(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.

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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 $\sin\left(\frac x8\right)$.

$$\begin{align*} \cos\left(\frac x8\right) &= \frac8\pi \sum_{n=1}^\infty \left(\frac{1-(-1)^n \cos\left(\frac\pi8\right)}{8n-1} + \frac{1-(-1)^n \cos\left(\frac\pi8\right)}{8n+1}\right) \sin(nx) \\[1ex] \sin\left(\frac x8\right) &= \frac{16}\pi \sin^2\left(\frac\pi{16}\right) - \frac8\pi \sum_{n=1}^\infty \left(\frac{1-(-1)^n \cos\left(\frac\pi8\right)}{8n-1} - \frac{1 - (-1)^n \cos\left(\frac\pi8\right)}{8n+1}\right) \cos(nx)\\[1ex] \sin\left(\frac x8\right)&= -\frac8\pi \sin\left(\frac\pi8\right) \sum_{n=1}^\infty \left(\frac{(-1)^n}{8n-1} + \frac{(-1)^n}{8n+1}\right) \sin(nx) \end{align*}$$

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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.

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I'm open to other methods for evaluating the series, but I'm more interested in the Fourier approach, if it's tenable.

Answer 1

Using the function suggested in comments (one of which contains a typo; $29$ should have been $28$), I arrive at the identity

$$\pi^2 \left(\left(x - \frac12\right)^2 - \frac{21}{256}\right) = \\ \cos(2\pi x) \sum_{k=0}^\infty \frac{(-1)^k}{(8k+1)^2} + \cos(4\pi x) \sum_{k=0}^\infty \frac{(-1)^k}{(8k+2)^2} + \cdots + \cos(14\pi x) \sum_{k=0}^\infty \frac{(-1)^k}{(8k+7)^2}$$

For $m\in\{1,2,\ldots,7\}$, let

$$S_m = \sum_{k=0}^\infty \frac1{(8k+m)^2} \text{ and } {S_m}' = \sum_{k=0}^\infty \frac{(-1)^k}{(8k+m)^2}$$

Evaluating both sides of the identity with $x\in\left\{\frac18,\frac38,\frac58,\frac78\right\}$ yields $S_4=\frac{\pi^2}{256}$, and more importantly the relation

$${S_1}'-{S_3}'-{S_5}'+{S_7}' = \boxed{\frac{\pi^2}{8\sqrt2}}$$

and from here it's easy to show that the left side is the same as $S_1-S_3-S_5+S_7$, the sum we want.

Answer 2

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 $S=\pi^2/(8\sqrt{2})$.

Answer 3

Here's another way to evaluate this series.

Since the series converges absolutely,

$$ \begin{aligned} S&=\sum_{n=0}^{\infty}\left(\frac{1}{(8n+1)^2}-\frac{1}{(8n+3)^2}-\frac{1}{(8n+5)^2}+\frac{1}{(8n+7)^2}\right) \\&=\frac{1}{64} \left(\psi ^{(1)}\left(\frac{1}{8}\right)-\psi ^{(1)}\left(\frac{3}{8}\right)-\psi ^{(1)}\left(\frac{5}{8}\right)+\psi ^{(1)}\left(\frac{7}{8}\right)\right) \end{aligned} $$.

With the identity of PolyGamma

$$ \psi ^{(1)}(1-x)+\psi ^{(1)}(x)=\pi ^2 \csc ^2(\pi x) $$

we get

$$ \begin{aligned}S&=\frac{\pi^2}{64}\left(\csc^2\left(\frac{\pi^2}{8}\right)-\csc^2\left(\frac{3\pi^2}{8}\right)\right) \\&=\frac{\pi^2}{64}\cdot 4 \sqrt{2} \\&=\frac{\sqrt{2}}{16}\pi^2 \end{aligned} $$

Answer 4

Just to clarify

$$\sum\limits_{n=0}^\infty\frac{(-1)^{n (n+1)/2}}{(2 n+1)^s}\tag{1}$$

is equivalent to

$$\sum\limits_{n=1}^{\infty }\frac{\chi_{8,2}(n)}{n^s}=L_{8,2}(s)=8^{-s} \left(\zeta \left(s,\frac{1}{8}\right)-\zeta \left(s,\frac{3}{8}\right)-\zeta \left(s,\frac{5}{8}\right)+\zeta \left(s,\frac{7}{8}\right)\right)\tag{2}$$

where $\chi_{8,2}(n)$ is the Dirichlet character $\{1,0,-1,0,-1,0,1,0\}$ and $L_{8,2}(s)$ is the associated Dirichlet series leading to

$$\sum\limits_{n=0}^\infty\frac{(-1)^{n (n+1)/2}}{(2 n+1)^2}=L_{8,2}(2)=\frac{\pi^2}{8 \sqrt{2}}.\tag{3}$$

Answer 5

I will provide a computation without the need for special functions. Let us define a complex valued function with the poles and residues necessary to reconstruct our sum of interest. We start with a composition of periodic functions

\begin{equation} f(z)= \underbrace{\frac{1}{\sin(8 \pi z)}}_{\substack{\text{provides} \\ \text{singularities}\\ \text{except}\,\,z=0} } \cdot \underbrace{\sin(4 \pi z)}_{\substack{\text{cancels} \\ \text{unnecessary} \\ \text{singularities}}} \cdot \underbrace{\sin(2 \pi z)}_{\substack{\text{adjusts} \\ \text{signs}\\ \text{of residues}}} \end{equation} now we combine this with the power law present in the target sum which also provides a singularity at $z=0$.

\begin{equation} g(z)=\frac{f(z)}{z^2} \end{equation} we have (left as an exercise to the reader, essentially this boils down to the fact that $|\sin(a z)| \sim e^{\Re(a z)} $ for large $z$): $|g(z)|\sim Const/|z^2|$ if $z \in \mathbb{C}/\{2 \mathbb{Z}+1 ,0\}$ and $|z| \rightarrow \infty$. As an integration contour we choose a large circle $C$ centered at zero such that it's boundary is always located between two singular points. In the limit of infinite circle radius we get by the residue theorem:

\begin{equation} 0=\oint_C g(z)dz = 2 \pi i \text{res}(g(z),z=0)+2 \pi i \sum_{m\in 2 \mathbb{Z}+1 } \text{res}(g(z),z=m/8)\quad \,\, (1) \end{equation} the residues of $g(z)$ correspond to a lot of first order singularities and can therefore calculated mechanically ($n\in \mathbb{Z}/0$): \begin{equation} \text{res}(g(z),z=m/8)=-\frac{4 \sqrt{2}}{\pi}\frac{a_m}{ m^2} \,\, \quad (2) \end{equation} \begin{equation} \text{res}(g(z),z=0)=\pi \quad \quad \quad \quad \quad \quad (3) \end{equation} with \begin{equation} a_m= \left\{\begin{array}{lr} +1, & \text{if } m \in \{8n+1,8n+7\} \\ -1, & \text{if } m \in \{8n+3,8n+5\} \end{array} \right\} \end{equation} we can now put (2) and (3) into (1) and obtain, using parity:

\begin{equation} \frac{\pi^2}{8 \sqrt{2}}=\sum_{m\in 2 \mathbb{N}+1 } \frac{a_m}{ m^2} \end{equation}

where the rhs is equal to the third representation of the sum in the OP