$\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.
 A: $\dfrac{1}{(3n+1)(3n+4)}=\dfrac{1}{3}\bigg(\dfrac{1}{3n+1}-\dfrac{1}{3n+4}\bigg)$
$\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}=\dfrac{1}{3}\bigg(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}\bigg)$ and so on ...
EDIT
$\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+...+\dfrac{1}{(3n+1)(3n+4)}=\dfrac{1}{3}\bigg(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{3n+1}-\dfrac{1}{3n+4}\bigg)=\dfrac{1}{3}\bigg(\dfrac{1}{4}-\dfrac{1}{3n+4}\bigg)$
So $\lim\limits_{n\to\infty}\bigg(\dfrac{1}{4\cdot 7}+\dfrac{1}{7\cdot 10}+...+\dfrac{1}{(3n+1)(3n+4)}\bigg)=\lim\limits_{n\to\infty}\dfrac{1}{3}\bigg(\dfrac{1}{4}-\dfrac{1}{3n+4}\bigg)=\dfrac{1}{12}$
A: By telescopic summation:
$$t_k=\frac{1}{3k+1}-\frac{1}{3k+4}=\frac{1}{3}\left(\frac{1}{3k+1}-\frac{1}{3k+4} \right)=[f_k-f_{k+1}]$$
So $$t_1=f_2-f_3, t_2=f_3-f_4, t_4=f_4-f_5, .....t_n=f_n-f_{n+1}$$
Adding them we have $$S_n=f_2-f_{n+1}=\frac{1}{3}[\frac{1}{4}-\frac{1}{3n+4}]$$
So, $$S_{\infty}=\frac{1}{12}$$
A: 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
$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac{1}{(3i+1)(3i+4)}$$
$$=\lim_{n\rightarrow\infty}\frac{1}{3}\big[\frac{1}{4}-\frac{1}{3n+4}\big]=\frac{1}{3}\cdot\frac{1}{4}=\frac{1}{12}.$$
A: $$S_n=\sum_{k=1}^{n} \frac{1}{(3k+1)(3k+4)}=\frac{1}{3}\sum_{k=1}^{n} \left(\frac{1}{3k+1}- \frac{1}{3k+3}\right)= \frac{1}{3} \sum_{k=0}^{n} \int_{1}^{1}[t^{3k}- t^{3k+3}] dt.$$
Use IGP: $$\sum_{k=1}^n r^k=\frac{r}{1-r}, |r|<1$$
$$S_{\infty}= \frac{1}{3} \int_{0}^{1} t^3\frac{1-t^3}{1-t^3} dt=\frac{1}{3}\int_{0}^{1} t^3 dt=\frac{1}{12}.$$
