I dont know how to solve this series with limit of partial sum.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(3n-2)(3n+1)}$
Thanx for any help
Hint: this is a telescoping sum
Note that $\displaystyle \frac 1{(3n - 2)(3n + 1)} = \frac 13 \cdot \frac {3n + 1 - (3n - 2)}{(3n - 2)(3n + 1)} = \frac 13 \left( \frac 1{3n - 2} - \frac 1 {3n + 1} \right)$
(instead of using algebraic manipulation as i did above, you can use partial fractions)