Proving convergence by the comparison test I need to prove the convergence of the following series :
$$\sum_{n=1}^{\infty} \frac{1}{(3n-2)(3n+1)}$$
I suppose I need to find two series like
$$\sum_{n=1}^{\infty}(k_1a_n + k_2b_n) = \sum_{n=1}^{\infty} = k_1\sum_{n=1}^{\infty}a_n  + k_2\sum_{n=1}^{\infty}b_n$$
I thought that the series $a_n$ and $b_n$ can be the partial sums, so
$$\sum_{n=1}^{\infty} \frac{1}{(3n-2)(3n+1)} =  \frac{1}{3}\sum_{n=1}^{\infty} \frac{1}{(3n+1)} - \frac{1}{3}\sum_{n=1}^{\infty} \frac{1}{(3n-2)}$$
I know, by the comparison test, that if $a_n$ and $b_n$ converge, the main series will it too. But, how can I demonstrate that the series $a_n$ and $b_n$ converge?

For $n\ge 1$, $3n+1>n$ and for $n\ge 2$, we have $3n-2>n$.  Hence,
$$\frac1{(3n-2)(3n+1)}<\frac1{n^2},\quad \text{for }n\ge 2.$$
