Determine whether the series is convergent or divergent by expressing $S_n$ as a telescoping sum $\sum_{n=1}^{\infty}\frac{6}{n(n+3)}$

Question

I have no idea where I'm going wrong or if I'm even doing this problem correctly. But here are my steps so far:

$$\sum_{n=1}^{\infty}\frac{6}{n(n+3)}=S_n\sum_{i=1}^{n}\frac{6}{i(i+3)}$$

After solving this:

$$\frac{6}{i(i+3)}=\frac{A}{i}+\frac{B}{i+3}$$

I got this to equal:

$$\sum_{i=1}^{n}\left(\frac{2}{i}-\frac{2}{i+3}\right)$$

So I wrote out the first 4 terms of this partial sum like this:

$$\left(\frac{2}{1}-\frac{2}{4}\right)+\left(\frac{2}{2}-\frac{2}{5}\right)+\left(\frac{2}{3}-\frac{2}{6}\right)+\left(\frac{2}{4}-\frac{2}{7}\right)+...+\left(\frac{2}{n}-\frac{2}{n+3}\right)$$

But I don't know if I did this right and I don't know what to do from here.

Answer 1

Hint: The answer is convergent. Write $\sum_{n=1}^{\infty}\frac{6}{n(n+3)}$ as

$$\sum_{n=1}^{\infty}\frac{6}{n(n+3)}=\sum_{k=1}^{\infty}\frac{6}{(3k-2)(3k+1)}+\sum_{k=1}^{\infty}\frac{6}{(3k-1)(3k+2)}+\sum_{k=1}^{\infty}\frac{6}{(3k)(3k+3)}$$

(WHY?)