<s>[INST] 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?
 [/INST] $$\sum_{n=1}^{\infty}\frac{3}{n(n+3)}=\lim_{k\to\infty}\sum_{n=1}^{k}\frac{3}{n(n+3)}=$$
$$=\lim_{k\to\infty}\left(\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)\right)=\frac{11}{6}$$
because
$$\sum_{n=1}^{k}\frac{3}{n(n+3)}=\sum_{n=1}^{k}\left(\frac{1}{n}-\frac{1}{n+3}\right)=\sum_{n=1}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+3}=$$
$$=1+\frac{1}{2}+\frac{1}{3}+\sum_{n=4}^{k}\frac{1}{n}-\sum_{n=1}^{k}\frac{1}{n+3}=\frac{11}{6}+\sum_{n=4}^{k}\frac{1}{n}-\sum_{j=4}^{k+3}\frac{1}{j}=$$
$$=\frac{11}{6}+\sum_{n=4}^{k}\frac{1}{n}-\left(\sum_{j=4}^{k}\frac{1}{j}+\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)=$$
$$=\frac{11}{6}-\left(\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}\right)$$
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