Đáp án:
$\displaystyle\sum\limits_{n = 1}^{\infty}\dfrac{1}{(3n+1)(3n+4)} =\dfrac{1}{12}$
Giải thích các bước giải:
$\quad \displaystyle\sum\limits_{n = 1}^{\infty}\dfrac{1}{(3n+1)(3n+4)}$
Tổng riêng thứ $n:$
$\quad S_n = \displaystyle\sum\limits_{i = 1}^n\dfrac{1}{(3i+1)(3i+4)}$
$\to S_n = \dfrac13\displaystyle\sum\limits_{i = 1}^n\dfrac{3}{(3i+1)(3i+4)}$
$\to S_n = \dfrac13\displaystyle\sum\limits_{i = 1}^n\left(\dfrac{1}{3i +1} -\dfrac{1}{3i+4}\right)$
$\to S_n = \dfrac13\left( \dfrac14 - \dfrac17+\dfrac17 - \dfrac{1}{10} + \cdots + \dfrac{1}{3n+1} - \dfrac{1}{3n+4}\right)$
$\to S_n = \dfrac13\left(\dfrac14 - \dfrac{1}{3n+4}\right)$
$\to \lim\limits_{n\to \infty}S_n = \lim\limits_{n\to \infty}\dfrac13\left(\dfrac14- \dfrac{1}{3n+4}\right)= \dfrac{1}{12}$
Vậy $\displaystyle\sum\limits_{n = 1}^{\infty}\dfrac{1}{(3n+1)(3n+4)} = \dfrac{1}{12}$
