The sequence $\left\{ a _ { n } \right\}$ is defined by

$$a _ { 1 } = \frac { 2 } { 9 } , \quad a _ { n } = \frac { ( n + 1 ) ( 2 n - 3 ) } { 3 n ( 2 n + 1 ) } a _ { n - 1 } \quad ( n = 2,3,4 , \cdots ) .$$

We are to find the general term $a _ { n }$ and the infinite sum $\sum _ { n = 1 } ^ { \infty } a _ { n }$.

(1) For A $\sim$ E in the following sentences, choose the correct answers from among (0) $\sim$ (9) below.

First, when we set $b _ { n } = \frac { n + 1 } { 3 ^ { n } a _ { n } }$ and express $\frac { b _ { n } } { b _ { n - 1 } }$ in terms of $n$, we have

$$\frac { b _ { n } } { b _ { n - 1 } } = \frac { \mathbf { A } } { \mathbf { B } } \cdot \frac { a _ { n - 1 } } { a _ { n } } = \frac { \mathbf { C } } { \mathbf { D } }$$

From this equation, we have

$$a _ { n } = \frac { n + 1 } { 3 ^ { n } ( \mathbf { E } ) ( 2 n + 1 ) } .$$

(0) $n - 1$\\
(1) $n$\\
(2) $n + 1$\\
(3) $2 n - 1$\\
(4) $2 n + 1$\\
(5) $2 n - 3$\\
(6) $2 n + 3$\\
(7) $3 n - 1$\\
(8) $3 n$\\
(9) $3 n + 1$\\
(2) Next, let $c _ { n } = \frac { 1 } { 3 ^ { n } ( 2 n + 1 ) } ( n = 0,1,2 , \cdots )$. When we set $a _ { n } = A c _ { n - 1 } + B c _ { n }$, we see that $A = \frac { \mathbf { F } } { \mathbf { G } }$ and $B = \frac { \mathbf { H I } } { \mathbf { G } }$. Using this result to find $S _ { n } = \sum _ { k = 1 } ^ { n } a _ { k }$, we have

$$S _ { n } = \frac { \mathbf { K } } { \mathbf { L } } \left( \mathbf { M } \right)$$

Hence we obtain

$$\sum _ { n = 1 } ^ { \infty } a _ { n } = \lim _ { n \rightarrow \infty } S _ { n } = \frac { \mathbf { N } } { \mathbf { O } }$$