Question

%7B%20-%201%29%7D%5E%7B1%20%2B%202%20%2B%203%20%2B%20%20%5Cdots%20%2B%20n%7D%20%7D%7B%282n%20%2B%201%20%7B%29%7D%5E%7B2%7D%20%7D" id="TexFormula1" title="\large \rm \sum \limits_{n = 0}^ \infty \frac{( { - 1)}^{1 + 2 + 3 + \dots + n} }{(2n + 1 {)}^{2} }" alt="\large \rm \sum \limits_{n = 0}^ \infty \frac{( { - 1)}^{1 + 2 + 3 + \dots + n} }{(2n + 1 {)}^{2} }" align="absmiddle" class="latex-formula">​

Answer 1

The sum we want is

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

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

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

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

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

$\displaystyle \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}$

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

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

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 $S'$-s :

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

which reduces to the identity

$\pi^2\left(\left(x-\dfrac12\right)^2-\dfrac{21}{256}\right) = \\\\ \cos(2\pi x) {S_1}' + \cos(4\pi x) {S_2}' + \cos(6\pi x) {S_3}' + \cos(8\pi x) {S_4}' \\\\ \,\,\,\, + \cos(10\pi x) {S_5}' + \cos(12\pi x) {S_6}' + \cos(14\pi x) {S_7}'$

Evaluating both sides at x for x ∈ {1/8, 3/8, 5/8, 7/8} and solving the system of equations yields the dependent solution

$\begin{cases}{S_4}' = \dfrac{\pi^2}{256} \\\\ {S_1}' - {S_3}' - {S_5}' + {S_7}' = \dfrac{\pi^2}{8\sqrt 2}\end{cases}$

It turns out that

${S_1}' - {S_3}' - {S_5}' + {S_7}' = S_1 - S_3 - S_5 + S_7$

so we're done, and the sum's value is $\boxed{\dfrac{\pi^2}{8\sqrt2}}$.