Answer
$\dfrac{1}{3}$
Work Step by Step
Here, we have the sum $S$ is the limit of the sequence of partial sums which is equal to:
$S_n=\Sigma_{k=0}^{n-1}\dfrac{1}{(3k+1)(3k+4)} =\dfrac{1}{3}\Sigma_{k=0}^{n-1} (\dfrac{1}{(3k+1)} -\dfrac{1}{(3k+4)} )$
Next, we have:
$S_n=\dfrac{1}{3}\Sigma_{k=0}^{n-1} (\dfrac{1}{(3k+1)} -\dfrac{1}{(3k+4)} )\\=\dfrac{1}{3} [(1-\dfrac{1}{4})+(\dfrac{1}{4}-\dfrac{1}{7})+......+(\dfrac{1}{3n-2}-\dfrac{1}{3n+1})]\\=\dfrac{1}{3}(1-\dfrac{1}{3n+1})$
Because for all $n \gt 0$ , $\lim\limits_{x \to \infty} x^{-n}=0$ and $\lim\limits_{x \to -\infty} x^{n}=\infty$
Thus, $S=\lim\limits_{n \to \infty} S_n=\dfrac{1}{3}(1-0)=\dfrac{1}{3}$
