\(\begin{array}{l}
\displaystyle\sum\limits_{n=1}^{\infty}\dfrac{1}{(3n+1)(3n+4)}\qquad\ (1)\\
\displaystyle\sum\limits_{n=1}^{\infty}\left(\sqrt2 + a \right)^{2n}\qquad\quad\qquad (2)\\
a)\quad \displaystyle\sum\limits_{n=1}^{\infty}\dfrac{1}{(3n+1)(3n+4)}\\
\text{Tổng riêng thứ n}\\
\quad S_n = \displaystyle\sum\limits_{i=1}^{n}\dfrac{1}{(3i+1)(3i+4)}\\
\Leftrightarrow S_n = \dfrac13\displaystyle\sum\limits_{i=1}^{n}\dfrac{3}{(3i+1)(3i+4)}\\
\Leftrightarrow S_n = \dfrac13\displaystyle\sum\limits_{i=1}^{n}\left(\dfrac{1}{3i+1} - \dfrac{1}{3i+4}\right)\\
\Leftrightarrow S_n = \dfrac13\left(\dfrac14 - \dfrac17 + \dfrac17 - \dfrac{1}{10} + \cdots + \dfrac{1}{3n+1} - \dfrac{1}{3n+4}\right)\\
\Leftrightarrow S_n = \dfrac13\left(\dfrac14 - \dfrac{1}{3n+4}\right)\\
\Leftrightarrow S_n = \dfrac{1}{12} - \dfrac{1}{3(3n+4)}\\
\text{Ta có:}\\
\lim\limits_{n\to\infty}S_n = \lim\limits_{n\to\infty}\left(\dfrac{1}{12} - \dfrac{1}{3(3n+4)}\right) = \dfrac{1}{12}\\
\text{Do đó chuỗi đã cho hội tụ}\\
b)\quad \displaystyle\sum\limits_{n=1}^{\infty}\left(\sqrt2 + a \right)^{2n}\\
+)\quad \text{Với $a = -\sqrt2 + \dfrac13$ ta được:}\\
\quad \displaystyle\sum\limits_{n=1}^{\infty}\left(\dfrac13\right)^{2n}\\
\Leftrightarrow \displaystyle\sum\limits_{n=1}^{\infty}\left(\dfrac19\right)^{n}\\
\text{Tổng riêng thứ n}\\
\quad S_n = \displaystyle\sum\limits_{i=1}^{n}\left(\dfrac19\right)^{i}\\
\Leftrightarrow S_n = \dfrac19 + \dfrac{1}{81} + \cdots + \left(\dfrac19\right)^{n}\\
\Leftrightarrow \dfrac19S_n = \dfrac{1}{81} + \dfrac{1}{729} + \cdots + \left(\dfrac19\right)^{n+1}\\
\Leftrightarrow \left(1 - \dfrac19\right)S_n = \dfrac19 - \left(\dfrac19\right)^{n+1}\\
\Leftrightarrow S_n = \dfrac{\dfrac19 - \left(\dfrac19\right)^{n+1}}{1 - \dfrac19}\\
\Leftrightarrow S_n = \dfrac18 - \dfrac18\cdot \left(\dfrac19\right)^{n}\\
\text{Ta có:}\\
\lim\limits_{n \to\infty}S_n = \lim\limits_{n \to\infty}\left[\dfrac18 - \dfrac18\cdot \left(\dfrac19\right)^{n}\right] = \dfrac18\\
\text{Do đó chuỗi số trên hội tụ}\\
\quad \displaystyle\sum\limits_{n=1}^{\infty}\left(\sqrt2 + a \right)^{2n}\\
\Leftrightarrow \displaystyle\sum\limits_{n=1}^{\infty}\left[\left(\sqrt2 + a \right)^{2}\right]^{n}\\
\text{Chuỗi số trên là chuỗi cấp số nhân với cơ số}\ q = \left(\sqrt2 + a \right)^{2}\\
\text{Chuỗi phân kỳ}\ \Leftrightarrow |q| \geqslant 1 \Leftrightarrow \left(\sqrt2 + a \right)^{2}\geqslant 1\\
\Leftrightarrow \left[\begin{array}{l}\sqrt2 + a \geqslant 1\\\sqrt2 + a \leqslant -1\end{array}\right.\ \Leftrightarrow \left[\begin{array}{l} a \geqslant 1 - \sqrt2\\a \leqslant -1 - \sqrt2\end{array}\right.
\end{array}\)