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?

One may just observe a telescoping sum here

$$
\sum_{n=1}^N \frac{1}{(3n-2)(3n+1)}=\frac13\sum_{n=1}^N \left(\frac{1}{3(n-1)+1}-\frac{1}{3n+1}\right)=\frac13-\frac{1}{3(3N+1)}
$$ 

giving the convergence and the sum of your initial series.
