Understanding the solution of a telescoping sum $\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$

I’m having trouble understanding infinite sequence and series as it relates to calculus, but I think I’m getting there.

For the below problem:

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

The solution shows them breaking this up into a sum of partial fractions. I understand how they got the first two terms, but then they show the partial fractions of the $n$ terms and I find myself lost.

Here is the what I’m talking about:

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

The next few terms are shown to be this:

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

And it continues but this is the part where I get confused…

$$\left(\frac{1}{n-3}-\frac{1}{n}\right)+\left(\frac{1}{n-2}-\frac{1}{n+1}\right)+\left(\frac{1}{n-1}-\frac{1}{n+2}\right)+\left(\frac{1}{n}-\frac{1}{n+3}\right)$$

Where did the $n$ terms in the denominator come from?

Solutions Collecting From Web of "Understanding the solution of a telescoping sum $\sum_{n=1}^{\infty}\frac{3}{n(n+3)}$"

$$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}=\lim_{k\to\infty}\sum_{n=1}^{k}\frac{3}{n(n+3)}=$$
$$=\lim_{k\to\infty}\left(\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)\right)=\frac{11}{6}$$
because
$$\sum_{n=1}^{k}\frac{3}{n(n+3)}=\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+3}\right)=\sum_{n=1}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+3}=$$
$$=1+\frac{1}{2}+\frac{1}{3}+\sum_{n=4}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+3}=\frac{11}{6}+\sum_{n=4}^{k}\frac{1}{n}-\sum_{j=4}^{k+3}\frac{1}{j}=$$
$$=\frac{11}{6}+\sum_{n=4}^{k}\frac{1}{n}-\left(\sum_{j=4}^{k}\frac{1}{j}+\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)=$$
$$=\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)$$