\(\begin{array}{l}
a)\quad \displaystyle\sum\limits_{n=1}^{\infty}\ln\left(1 - \dfrac{2}{2n+1}\right)\qquad (1)\\
\text{Tổng riêng thứ n}\\
\quad S_n = \displaystyle\sum\limits_{i=1}^{n}\ln\left(1 - \dfrac{2}{2i+1}\right)\\
\Leftrightarrow S_n = \displaystyle\sum\limits_{i=1}^{n}\ln\left(\dfrac{2i-1}{2i+1}\right)\\
\Leftrightarrow S_n = \ln\dfrac13 + \ln\dfrac35 + \ln\dfrac57 + \cdots + \ln\dfrac{2n-1}{2n+1}\\
\Leftrightarrow S_n = \ln\left(\dfrac13\cdot \dfrac35\cdot \dfrac57\cdots\dfrac{2n-1}{2n+1}\right)\\
\Leftrightarrow S_n = \ln\dfrac{1}{2n+1}\\
\text{Ta có:}\\
\quad \lim\limits_{n\to \infty}S_n = \lim\limits_{n\to \infty}\ln\dfrac{1}{2n+1} = -\infty\\
\text{Do đó chuỗi đã cho phân kỳ}
\end{array}\)
\(\begin{array}{l}b)\quad \displaystyle\sum\limits_{n=1}^{\infty}\left(3+a\right)^{2n}\qquad (2)\\\text{Với $a=\dfrac{\sqrt5}{2} - 3$ ta được:}\\\quad \displaystyle\sum\limits_{n=1}^{\infty}\left(\dfrac{\sqrt5}{2}\right)^{2n}\\= \displaystyle\sum\limits_{n=1}^{\infty}\left(\dfrac54\right)^{n}\\\text{Tổng riêng thứ n:}\\\quad S_n =\displaystyle\sum\limits_{i=1}^{n}\left(\dfrac54\right)^{i}\\\Leftrightarrow S_n = \dfrac54 + \dfrac{25}{16} +\cdots + \left(\dfrac54\right)^{n}\\\Leftrightarrow \dfrac54S_n = \dfrac{25}{16} + \dfrac{125}{64} +\cdots + \left(\dfrac54\right)^{n+1}\\\Leftrightarrow \left(1-\dfrac54\right)S_n = \dfrac54 - \left(\dfrac54\right)^{n+1}\\\Leftrightarrow S_n = 4\cdot \left(\dfrac54\right)^{n+1}-5\\\text{Ta có:}\\\lim\limits_{n\to \infty}S_n = \lim\limits_{n\to \infty}\left[\left(\dfrac54\right)^{n+1}-5\right] = +\infty\\\text{Do đó chuỗi trên phân kỳ}\\\text{Xét}\ \ \displaystyle\sum\limits_{n=1}^{\infty}\left(3+a\right)^{2n}\\=\displaystyle\sum\limits_{n=1}^{\infty}\left[(3+a)^2\right]^{n}\\\text{Chuỗi $(2)$ là chuối cấp số nhân với công bội}\ |q| = (3 + a)^2\\\text{Chuỗi $(2)$ phân kỳ}\ \Leftrightarrow |q|\geqslant 1\\\Leftrightarrow (3+a)^2 \geqslant 1\\\Leftrightarrow \left[\begin{array}{l}3 + a \geqslant 1\\3 + a \leqslant -1\end{array}\right.\\\Leftrightarrow \left[\begin{array}{l}a \geqslant -2\\a \leqslant - 4\end{array}\right.\end{array}\)
