What is the limit of this specific series?

Question

Anyone can give me the path for this, could not figure out which theory/method to use for this... $$\lim_{n\to\infty}\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7} + \frac{1}{7\cdot10}+.....+ \frac {1}{(3n-2)\cdot(3n+1)}\right)=?$$

Answer 1

Hint: $\displaystyle\frac1{1\times4}=\frac13\left(1-\frac14\right)$, $\displaystyle\frac1{4\times7}=\frac13\left(\frac14-\frac17\right)$, $\displaystyle\frac1{7\times10}=\frac13\left(\frac17-\frac1{10}\right)$, …

Answer 2

If you break the terms in this way $$\frac{1}{(3n-2)(3n+1)}=\frac{1/3}{3n-2}-\frac{1/3}{3n+1}$$

many will cancel.

Answer 3

HINT: It is $$ \lim_{n\to\infty}\frac13\left( \frac11-\frac14+\frac14-\frac17+\frac17-\frac1{10}\ldots+\frac1{3n-2}-\frac1{3n+1} \right) $$