Question

Sabendo que
$\begin{array}{l}\displaystyle\sum_{\sf n\:\geq\:1}\sf\dfrac{1}{~n^2}=\dfrac{1}{~1^2}+\dfrac{1}{~2^2}+\dfrac{1}{~3^2}+\:\dots\:=\dfrac{\pi}{6}\end{array}$
, então
$\begin{array}{l}\displaystyle\sum_{\sf n\:\geq\:1}\sf \dfrac{1}{~n^2(n+1)^2}=\dfrac{1}{~1^22^2}+\dfrac{1}{~2^23^2}+\dfrac{1}{~3^24^2}+\:\dots\:\end{array}$
é igual a:
⠀
A) $\begin{array}{l}\sf\dfrac{~\pi^2}{6}-1\end{array}$
⠀
B) $\begin{array}{l}\sf\dfrac{~\pi^2}{6}\bigg(\dfrac{~\pi^2}{6}-1\bigg)\end{array}$
⠀
C)$\begin{array}{l}\sf\dfrac{~\pi^2}{3}-3\end{array}$
⠀
D) $\begin{array}{l}\sf\dfrac{~\pi^2}{3}+1\end{array}$
⠀
E) $\begin{array}{l}\sf\dfrac{~\pi^4}{9}-2\end{array}$

Answer 1

Boa Tarde!

$\mathrm {\displaystyle \sum_{n \geq 1} \dfrac {1}{n^2(n+1)^2} = \displaystyle \sum_{n\geq 1} \dfrac{(n+1)^2+ n^2- 2n (n+1)}{n^2(n+1)^2} = \displaystyle \sum_{n\geq 1}\dfrac{1}{n^2} + \displaystyle \sum_{n\geq 1} \dfrac{1}{(n+1)^2} -2 \displaystyle \sum_{n\geq 1} \dfrac{1}{n(n+1)} }$$\mathrm { \displaystyle \sum_{n\geq 1} \dfrac{1}{n^2} = \dfrac{\pi ^2}{6} \: e \: \displaystyle \sum_{n\geq 1} \dfrac{1}{(n+1)^2} = \boxed {\dfrac{\pi ^2}{6} -1} }$

$\mathrm {\displaystyle \sum_{n\geq 1} \dfrac{1}{n(n+1)} = \displaystyle \sum_{n\geq 1} \left ( \dfrac{1}{n} - \dfrac{1}{n+1} \right ) = \boxed {1} }$

$\mathrm {Portanto: } \\ \\ \mathrm { \dfrac{\pi ^2}{6} + \dfrac{\pi ^2}{6} -1-2 = \boxed {\boxed { \dfrac{\pi ^2}{3} -3 } } }$

Letra C)

• Att. MatiasHP

Answer 2

Resposta: Letra C)

Reescrevendo o segundo somatório e chamando-o de S, obtemos:

$\tt \displaystyle S=\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2(n+1)^2}=\sum_{n\,=\,1}^{\infty}\left(\dfrac{1}{n(n+1)}\right)^{\!\!2}=\sum_{n\,=\,1}^{\infty}\left(\dfrac{(n+1)-n}{n(n+1)}\right)^{\!\!2}\\\\\\ \tt S=\sum_{n\,=\,1}^{\infty}\left(\dfrac{n+1}{n(n+1)}-\dfrac{n}{n(n+1)}\right)^{\!\!2}=\sum_{n\,=\,1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)^{\!\!2}\qquad (\:1\:)$

Em seguida, vamos aplicar o quadrado da diferença de dois termos na expressão ( 1 ):

$\tt S=\displaystyle\sum_{n\,=\,1}^{\infty}\left[\left(\dfrac{1}{n}\right)^{\!\!2}-2\cdot \dfrac{1}{n}\cdot \dfrac{1}{n+1}+\left(\dfrac{1}{n+1}\right)^{\!\!2}\right]\\\\\\ \tt S=\sum_{n\,=\,1}^{\infty}\left(\dfrac{1}{n^2}-\dfrac{2}{n(n+1)}+\dfrac{1}{(n+1)^2}\right)\qquad (\:2\:)$

Aplicando uma das propriedades do somatório na expressão ( 2 ), encontraremos:

$\tt S=\displaystyle\sum_{i\,=\,1}^{\infty}\dfrac{1}{n^2}-\sum_{n\,=\,1}^{\infty}\dfrac{2}{n(n+1)}+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}\\\\\\ \tt S=\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}-2\cdot\! \sum_{n\,=\,1}^{\infty}\dfrac{1}{n(n+1)}+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}$

$\tt S=\displaystyle\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}-2\:\!\cdot\!\!\:\!\sum_{n\,=\,1}^{\infty}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}\\\\\\ \tt S=\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}-2\:\!\cdot \!\lim_{n\to\infty}\left[\,\sum_{k\,=\,1}^{n}\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)\right]+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}$

$\tt S=\displaystyle\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}-2\:\!\cdot\! \lim_{n\to\infty}\left(1-\dfrac{1}{n+1}\right)+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}\\\\\\ \tt S=\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}-2\cdot\! \lim_{n\to \infty}\left(\dfrac{n}{n+1}\right)+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}\\\\\\ \tt S=\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}-2\cdot\! \lim_{n\to \infty}\left[\dfrac{\diagup\!\!\!\!n}{\diagup\!\!\!\!n\left(1+\dfrac{1}{n}\right)}\right]+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}$

$\tt S=\displaystyle\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}-2\cdot \!\lim_{n\to \infty}\left[\dfrac{1}{1+\dfrac{1}{n}}\right]+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}\\\\\\ \tt S=\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}-2\cdot \dfrac{\displaystyle\lim_{n\to\infty}1}{\displaystyle\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)}+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}$

$\tt S=\displaystyle\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}-2\:\!\cdot \dfrac{\displaystyle\lim_{n\to\infty}1}{\displaystyle\lim_{n\to\infty}1\:\!+\lim_{n\to\infty}\left(\dfrac{1}{n}\right)}+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}\\\\\\ \tt S=\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}-2\:\!\cdot \dfrac{1}{1+0}+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}\\\\\\ \tt S=\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}\:\!+\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}-2\qquad(\:3\:)$

Do enunciado, inferimos que:

$\tt \!\bullet\ \ \displaystyle\sum_{n\,=\,1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}\qquad Obs.\!\!\:\!:\, o\ correto\ \'e\ \, \dfrac{\pi^2}{6}\qquad (\:4\:)\\\\\\ \tt \bullet\ \ \displaystyle\sum_{n\,=\,1}^{\infty}\dfrac{1}{(n+1)^2}=\sum_{n\,=\,1}^{\infty}\left(\dfrac{1}{n^2}\right)\!\:\!\:\!-1=\dfrac{\pi^2}{6}-1\qquad(\:5\:)$

Para finalizar, vamos substituir ( 4 ) e ( 5 ) em ( 3 ) e achar o valor de S:

$\tt S=\dfrac{\pi^2}{6}+\left(\dfrac{\pi^2}{6}-1\right)-2\\\\\\ \tt S=\dfrac{\pi^2}{6}+\dfrac{\pi^2}{6}-1-2\\\\\\ \tt S=\diagup\!\!\!\!2\cdot \dfrac{\pi^2}{\diagup\!\!\!\!2\cdot 3}-3\\\\\\ \boxed{\tt S=\dfrac{\pi^2}{3}-3}$