$a)\quad \displaystyle\sum\limits_{n=1}^{\infty}\dfrac{1}{n^2 + n}\qquad (1)$

Tổng riêng thứ $n:$

$\quad S_n = \displaystyle\sum\limits_{i=1}^{n}\dfrac{1}{i^2 + i}$

$\Leftrightarrow S_n = \displaystyle\sum\limits_{i=1}^{n}\dfrac{i+1 - i}{i(i+1)}$

$\Leftrightarrow S_n = \displaystyle\sum\limits_{i=1}^{n}\left(\dfrac{1}{i} - \dfrac{1}{i+1}\right)$

$\Leftrightarrow S_n = 1 - \dfrac12 + \dfrac12 - \dfrac13 +\cdots + \dfrac{1}{n} - \dfrac{1}{n+1}$

$\Leftrightarrow S_n = 1 -\dfrac{1}{n+1}$

Ta có:

$\lim\limits_{n\to \infty}S_n = \lim\limits_{n\to \infty}\left(1 -\dfrac{1}{n+1}\right) = 1$

Do đó chuỗi đã cho hội tụ

$b)\quad \displaystyle\sum\limits_{n = 1}^{\infty}(a-2)^{2n}\qquad (2)$

Với $a = \dfrac{\sqrt6}{2} - 2$ ta được:

$\quad \displaystyle\sum\limits_{n = 1}^{\infty}\left(\dfrac{\sqrt6}{2}\right)^{2n}$

$= \displaystyle\sum\limits_{n = 1}^{\infty}\left(\dfrac{3}{2}\right)^{n}$

Tổng riêng thứ $n:$

$\quad S_n = \displaystyle\sum\limits_{i = 1}^{n}\left(\dfrac{3}{2}\right)^{i}$

$\Leftrightarrow S_n = \dfrac32 + \dfrac94 + \cdots +\left(\dfrac{3}{2}\right)^{n}$

$\Leftrightarrow \dfrac32S_n = \dfrac94 + \dfrac{81}{16} + \cdots + \left(\dfrac{3}{2}\right)^{n+1}$

$\Leftrightarrow \left(1-\dfrac32\right)S_n = \dfrac32 - \left(\dfrac{3}{2}\right)^{n+1}$

$\Leftrightarrow S_n = 2\cdot \left(\dfrac{3}{2}\right)^{n+1} - 3$

Ta có:

$\lim\limits_{n\to \infty}S_n = \lim\limits_{n\to \infty}\left[2\cdot \left(\dfrac{3}{2}\right)^{n+1} - 3\right]= +\infty$

Do đó chuỗi trên phân kỳ

Xét $\displaystyle\sum\limits_{n = 1}^{\infty}(a-2)^{2n}$

$= \displaystyle\sum\limits_{n = 1}^{\infty}[(a-2)^{2}]^{n}$

Chuỗi đã cho là chuỗi cấp số nhân với công bội $|q|= (a-2)^2$

Chuỗi phân kỳ $\Leftrightarrow |q|\geqslant 1$

$\Leftrightarrow (a-2)^2 \geqslant 1$

$\Leftrightarrow \left[\begin{array}{l}a-2\geqslant 1\\a - 2 \leqslant -1\end{array}\right.$

$\Leftrightarrow \left[\begin{array}{l}a\geqslant 3\\a \leqslant 1\end{array}\right.$