Q:

Calculate $\lim\limits_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{6}{k(k+1)(k+3)}$

$$\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{6}{k(k+1)(k+3)}$$
I tried to simplify the sum and I got $\frac{2}{k}-\frac{3}{k+1}+\frac{1}{k+3}$ but  I can't use this to simplify the terms.Also,I tried to amplify with $k+2$ and I got $$\frac{(k+3)-1}{k(k+1)(k+2)(k+3)}=\frac{(k+3)}{k(k+1)(k+2)(k+3)}-\frac{1}{k(k+1)(k+2)(k+3)}$$ but the terms also don't simplify.

A:

Hint
$$\frac{2}{k}-\frac{3}{k+1}+\frac{1}{k+3}=\frac{2}{k}-\frac{2}{k+1}-\frac{1}{k+1}+\frac{1}{k+3}=2\left(\frac{1}{k}-\frac{1}{k+1} \right)-\left(\frac{1}{k+1}-\frac{1}{k+3} \right)$$

