<s>[INST] $\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right) $ I am having trouble finding the infinite sum
$$
\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right).
$$
I know that
$$
\lim_{n\to\infty}\frac1{(3n+1)(3n+4)} =0,
$$
but I have no ideas for a further solution.
 [/INST] You want to compute $$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac{1}{(3i+1)(3i+4)}.$$
Using partial fractions we want to find constants $A$ and $B$ with $$\frac{1}{(3n+1)(3n+4)}=\frac{A}{3n+1}+\frac{B}{3n+4}=\frac{A(3n+4)+B(3n+1)}{(3n+1)(3n+4)}$$
So comparing numerators we want $1=A(3n+4)+B(3n+1)$. Then choosing $n=-\frac{1}{3}$ we obtain $1=3A$ so $A=\frac{1}{3}$ and similarly $B=-\frac{1}{3}$.
We then have $$\frac{1}{(3n+1)(3n+4)}=\frac{1}{3}\big(\frac{1}{3n+1}-\frac{1}{3n+4}\big)$$
So lets see what happens to the partial sum $$\sum_{i=1}^{n}\frac{1}{(3i+1)(3i+4)}=\frac{1}{3}\sum_{i=0}^{n}\big[\frac{1}{3i+1}-\frac{1}{3i+4}\big]$$
$$=\frac{1}{3}\big[(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})+\frac{1}{10}+...+(\frac{1}{3n-2}-\frac{1}{3n+1})+(\frac{1}{3n+1}-\frac{1}{3n+4})\big]$$
$$=\frac{1}{3}\big[\frac{1}{4}-\frac{1}{3n+4}\big]$$
Thus taking the limit we obtain
$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac{1}{(3i+1)(3i+4)}$$
$$=\lim_{n\rightarrow\infty}\frac{1}{3}\big[\frac{1}{4}-\frac{1}{3n+4}\big]=\frac{1}{3}\cdot\frac{1}{4}=\frac{1}{12}.$$
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