1 Answer

Rosalee · Answer 1 · 2023-07-19T20:23:24+0000

The sum we want is

$\displaystyle \sum_(n=0)^\infty ((-1)^(T_n))/((2n+1)^2) = 1 - \frac1{3^2} - \frac1{5^2} + \frac1{7^2} + \cdots$

where
$T_n=\frac{n(n+1)}2$ is the n-th triangular number, with a repeating sign pattern (+, -, -, +). We can rewrite this sum as

$\displaystyle \sum_(k=0)^\infty \left(\frac1{(8k+1)^2} - \frac1{(8k+3)^2} - \frac1{(8k+7)^2} + \frac1{(8k+7)^2}\right)$

For convenience, I'll use the abbreviations

$S_m = \displaystyle \sum_(k=0)^\infty \frac1{(8k+m)^2}$

${S_m}' = \displaystyle \sum_(k=0)^\infty ((-1)^k)/((8k+m)^2)$

for m ∈ {1, 2, 3, …, 7}, as well as the well-known series

$\displaystyle \sum_(k=1)^\infty ((-1)^k)/(k^2) = -(\pi^2)/(12)$

We want to find
S_1-S_3-S_5+S_7 .

Consider the periodic function
$f(x) = \left(x-\frac12\right)^2$ on the interval [0, 1], which has the Fourier expansion

$f(x) = \frac1{12} + \frac1{\pi^2} \sum_(n=1)^\infty (\cos(2\pi nx))/(n^2)$

That is, since f(x) is even,

$f(x) = a_0 + \displaystyle \sum_(n=1)^\infty a_n \cos(2\pi nx)$

where

$a_0 = \displaystyle \int_0^1 f(x) \, dx = \frac1{12}$

$a_n = \displaystyle 2 \int_0^1 f(x) \cos(2\pi nx) \, dx = \frac1{n^2\pi^2}$

(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
-s :

$\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_(k=0)^\infty (\cos(2\pi(8k+1) x))/((8k+1)^2) + \sum_(k=0)^\infty (\cos(2\pi(8k+2) x))/((8k+2)^2) + \cdots \right. \\ \,\,\,\, \left. + \sum_(k=0)^\infty (\cos(2\pi(8k+7) x))/((8k+7)^2) + \sum_(k=1)^\infty (\cos(2\pi(8k) x))/((8k)^2)\right)$

which reduces to the identity

$\pi^2\left(\left(x-\frac12\right)^2-(21)/(256)\right) = \\\\ \cos(2\pi x) {S_1}' + \cos(4\pi x) {S_2}' + \cos(6\pi x) {S_3}' + \cos(8\pi x) {S_4}' \\\\ \,\,\,\, + \cos(10\pi x) {S_5}' + \cos(12\pi x) {S_6}' + \cos(14\pi x) {S_7}'$