Demostrando que $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}$

Question

Quiero demostrar que la $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}, $$ el uso de la Serie de Fourier para la $2\pi$-función periódica $f(\theta)=\theta^{2},\quad (-\pi<0<\pi)$, es decir,

$$\theta^{2}=\frac{\pi^{2}}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(n\theta)}{n^{2}} $$ Así,

$$\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(n\theta)}{n^{2}}=\frac{\theta^{2}}{4}-\frac{\pi^{2}}{12}.$$

Estoy buscando un $\theta$ tal que $(-1)^{n}\cos(n\theta)=(-1)^{n+1}\Rightarrow \cos(n\theta)=-1$ para todos los $n\in\mathbb{N}$. ¿Cómo puedo elegir que $\theta$?

Answer 1

ps

Toma $$\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(n\theta)}{n^{2}}=\frac{\theta^{2}}{4}-\frac{\pi^{2}}{12}$ para $\theta=0$

$$ \ sum_ {n = 1} ^ {\ infty} \ frac {(- 1) ^ {n}} {n ^ {2}} = - \ frac {\ pi ^ {2}} {12} \\ (-1) \ sum_ {n = 1} ^ {\ infty} \ frac {(- 1) ^ {n}} {n ^ {2}} = (- 1) \ cdot \ left (- \ frac {\ pi ^ {2}} {12} \ right) \\ \ sum_ {n = 1} ^ {\ infty} \ frac {(- 1) ^ {n +1}} {n ^ {2}} = \ frac {\ pi ^ {2}} {12} \\ $$

Answer 2

Split! $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\sum_{n ~\text{odd}}\frac{1}{n^{2}} - \sum_{n ~\text{even}}\frac{1}{n^{2}}.$$

Sumar y restar el "incluso" parte:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}=\left(\sum_{n ~\text{impar}}\frac{1}{n^{2}} + \sum_{n ~\text{incluso}}\frac{1}{n^{2}}\right) - \sum_{n ~\text{incluso}}\frac{1}{n^{2}} - \sum_{n ~\text{incluso}}\frac{1}{n^{2}} = \\ =\sum_{n=1}^{\infty}\frac{1}{n^2}-2\sum_{n ~\text{incluso}}\frac{1}{n^{2}} = \frac{\pi^2}{6} - 2\sum_{n ~\text{incluso}}\frac{1}{n^{2}}.$$

Ahora, observe que:

$$\sum_{n ~\text{even}}\frac{1}{n^{2}}=\sum_{i =1}^{\infty}\frac{1}{(2i)^{2}} = \frac{1}{4}\sum_{i =1}^{\infty}\frac{1}{i^{2}} = \frac{1}{4}\frac{\pi^2}{6}.$$

Por lo tanto:

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}} = \frac{\pi^2}{6} - 2\frac{1}{4}\frac{\pi^2}{6} = \frac{\pi^2}{12}.$$