Calcular

Question

¿Cómo averiguo el valor de la siguiente serie infinita?

<span class="math-container">$$\sum_{n=0}^\infty \frac{1}{(4n^2-1)^2} $$</span>

Mi intento de una solución:

<span class="math-container">$$\sum{n=0}^\infty \frac{1}{(4n^2-1)^2} = \sum{n=0}^\infty \frac{1}{((2n-1)(2n+1))^2} = \sum{n=0}^\infty \left(\frac{1}{2}\left(\frac{1}{(2n-1)}-\frac{1}{(2n+1)}\right)\right)^2 = \frac{1}{4}\sum{n=0}^\infty \left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)^2 $$</span>

Entonces traté de calcular las sumas parciales de esta serie, pero sin suerte. ¿Alguien sabe cómo hacerlo?

Answer 1

Vamos a continuar con el enfoque que se tomó. Tenemos <span class="math-container">$$\begin{align}\sum{n=0}^\infty \frac{1}{(4n^2-1)^2} &=\frac{1}{4}\sum{n=0}^\infty \bigg(\frac{1}{2n-1}-\frac{1}{2n+1}\bigg)^2\ &=\frac{1}{4}\sum_{n=0}^\infty \frac{1}{(2n-1)^2}+\frac{1}{(2n+1)^2}-\frac{2}{(2n-1)(2n+1)} \end {Alinee el} $</span>

Ahora, usando el hecho de <span class="math-container">$$\sum{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{3}{4}\zeta(2)=\frac{\pi^2}{8}$ $</span> tenemos <span class="math-container">$$\sum{n=0}^\infty \frac{1}{(4n^2-1)^2}=\frac{3\zeta(2)+2}{8}-\frac{1}{2}\sum{n=0}^\infty \frac{1}{(2n-1)(2n+1)}$ $</span> ahora podemos utilizar telescópico para evaluar el último serie <span class="math-container">$$\frac{1}{2}\sum{n=0}^\infty \frac{1}{(2n-1)(2n+1)}=\frac{1}{4}\sum{n=0}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}=-\frac{1}{4}$</span> $ para que tengamos <span class="math-container">$$\sum{n=0}^\infty \frac{1}{(4n^2-1)^2}=\frac{3\zeta(2)+4}{8}=\frac{1}{2}+\frac{\pi^2}{16}$ $</span>

Answer 2

Tenga en cuenta<span class="math-container">$$(\forall n\in\mathbb{N}):\frac1{(4n^2-1)^2}=\frac1{4(2n+1)}-\frac1{4 (2n-1)}+\frac1{4(2n+1)^2}+\frac1{4(2n-1)^2}.$$It is clear that the series<span class="math-container">$$\sum{n=0}^\infty\frac1{4(2n+1)}-\frac1{4(2n-1)}$$</span>is a telescoping series, whose sum is <span class="math-container">$\frac14$</span>. On the other hand<span class="math-container">\begin{align}\sum{n=0}^\infty\frac{1}{4 (2 n+1)^2}+\frac{1}{4(2 n-1)^2}&=\frac14+2\sum{n=0}^\infty\frac{1}{4 (2 n+1)^2}\&=\frac14+\frac12\sum{n=0}^\infty\frac1{(2n+1)^2}.\end{align}</span>But, since<span class="math-container">$$\sum_{n=0}^\infty\frac1{(2n+1)^2}=\frac{\pi^2}8,$$</span>we have that the sum of your series is<span class="math-container">$$\frac14+\frac14+\frac{\pi^2}{16}=\frac12+\frac{\pi^2}{16}.$$</span></span>

Answer 3

Desde <span class="math-container">$$ \left|\sin x\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum{n\geq 1}\frac{\cos(2nx)}{4n^2-1} $ $</span> por Teorema de Parseval tenemos <span class="math-container">$$ \pi=\int{-\pi}^{\pi}\left|\sin x\right|^2\,dx =2\pi\cdot\frac{4}{\pi^2}+\frac{16}{\pi}\sum_{n\geq 1}\frac{1}{(4n^2-1)^2}$ $</span> por lo tanto, de reorganizar: <span class="math-container">%#% $ #%</span>