Question

Valor da soma desta seguinte serie:
$\sum_{n=1}^{\infty } \frac{-5}{9n^2+3n-2}$

Answer 1

Vamos encontrar as raízes de $p(x)=9x^{2}+3x-2$

$\Delta=3^{2}-4\cdot9\cdot(-2)\\\Delta=9+72\\\Delta=81\\\\x=\dfrac{-3\pm\sqrt{81}}{2\cdot9}=\dfrac{-3\pm9}{2\cdot9}=\dfrac{-1\pm3}{6}$

Então, as raízes de p(x) são

$x_{1}=\dfrac{-1+3}{6}=\dfrac{1}{3}\\\\\\x_{2}=\dfrac{-1-3}{6}=-\dfrac{2}{3}$

Com isso, podemos escrever p(x) como

$p(x)=9x^{2}+3x-2=9(x-\frac{1}{3})(x+\frac{2}{3})\\\\p(x)=9(\frac{3x-1}{3})(\frac{3x+2}{3})\\\\\boxed{\boxed{p(x)=(3x-1)(3x+2)}}$

Agora, nosso objetivo é encontrar números A e B tais que

$\dfrac{1}{9x^{2}+3x-2}=\dfrac{A}{3x-1}+\dfrac{B}{3x+2}\\\\\\\dfrac{1}{9x^{2}+3x-2}=\dfrac{A(3x+2)+B(3x-1)}{(3x-1)(3x+2)}\\\\\\\dfrac{1}{9x^{2}+3x-2}=\dfrac{3Ax+2A+3Bx-B}{9x^{2}+3x-2}\\\\\\\dfrac{1}{9x^{2}+3x-2}=\dfrac{(3A+3B)x+(2A-B)}{9x^{2}+3x-2}$

Isso ocorrerá se

$\begin{cases}3A+3B=0~~\therefore~~A+B=0\\2A-B=1\end{cases}$

Somando as equações, temos

$A+2A+B-B=0+1\\\\3A=1~~\therefore~~\boxed{\boxed{A=\dfrac{1}{3}}}$

Encontrando B:

$A+B=0~~\therefore~~~B=-A~~\therefore~~\boxed{\boxed{B=-\dfrac{1}{3}}}$

Então:

$\dfrac{1}{9n^{2}+3n-2}=\dfrac{(\frac{1}{3})}{3n-1}-\dfrac{(\frac{1}{3})}{3n+2}$

Logo:

$\displaystyle\sum\limits_{n=1}^{k}\dfrac{-5}{9n^{2}+3n-2}=-5\sum\limits_{n=1}^{k}\dfrac{1}{9n^{2}+3n-2}\\\\\\\sum\limits_{n=1}^{k}\dfrac{-5}{9n^{2}+3n-2}=-5\sum\limits_{n=1}^{k}\bigg[\dfrac{(\frac{1}{3})}{3n-1}-\dfrac{(\frac{1}{3})}{3n+2}\bigg]\\\\\\\boxed{\boxed{\sum\limits_{n=1}^{k}\dfrac{-5}{9n^{2}+3n-2}=-\dfrac{5}{3}\sum\limits_{n=1}^{k}\bigg[\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\bigg]}}$
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Expandindo a soma $\sum\limits_{n=1}^{k}\bigg[\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\bigg]}}$, temos

$S_{k}=\sum\limits_{n=1}^{k}\bigg[\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\bigg]}}=\\\\\\(\frac{1}{2}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11})+..+(\frac{1}{3k-4}-\frac{1}{3k-1})+(\frac{1}{3k-1}-\frac{1}{3k+2})$

Note que todos os termos se cancelam, com exceção do primeiro e do último:

$S_{k}=\dfrac{1}{2}-\dfrac{1}{3k+2}$

Finalmente, temos, por definição, que

$\displaystyle\sum\limits_{n=1}^{\infty}a_{n}=\lim\limits_{k\rightarrow\infty}\sum\limits_{n=1}^{k}a_{n}$

se o limite existir. Então:

$\displaystyle\sum\limits_{n=1}^{\infty}\dfrac{-5}{9n^{2}+3n-2}=\lim\limits_{k\rightarrow\infty}-\dfrac{5}{3}\bigg[\dfrac{1}{2}-\dfrac{1}{3k+2}\bigg]$

Quando $n\rightarrow\infty$, $3k+2\rightarrow\infty~~\Rightarrow~~\frac{1}{3k+2}\rightarrow0$

$\displaystyle\sum\limits_{n=1}^{\infty}\dfrac{-5}{9n^{2}+3n-2}=\lim\limits_{k\rightarrow\infty}-\dfrac{5}{3}\bigg[\dfrac{1}{2}-\dfrac{1}{3k+2}\bigg]\\\\\\\sum\limits_{n=1}^{\infty}\dfrac{-5}{9n^{2}+3n-2}=-\dfrac{5}{3}\bigg[\dfrac{1}{2}-0\bigg]\\\\\\\sum\limits_{n=1}^{\infty}\dfrac{-5}{9n^{2}+3n-2}=-\dfrac{5}{3}\cdot\dfrac{1}{2}\\\\\\\boxed{\boxed{\sum\limits_{n=1}^{\infty}\dfrac{-5}{9n^{2}+3n-2}=-\dfrac{5}{6}}}$