Telescoping Series confusion My professor gave us this example on her notes:
$$\sum_{n = 1}^\infty   \left(\frac{3}{n(n+3)}+\frac{1}{2^n}\right)$$
So I know we're supposed to find the partial fraction, which ends up being
$$\left(\frac{3}{n(n+3)}=\frac{A}{n}+\frac{B}{n+3}=
\frac{1}{n}-\frac{1}{n+3}\right)$$
So based on how she did the other examples, I would expect her to do:
$$\sum_{n = 1}^\infty   \left(\frac{3}{n(n+3)}=\frac{1}{1}-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}\right.....)$$, because I'd be plugging in numbers for n starting with n=1. However, she instead did the following:
$$\sum_{n = 1}^\infty   \left(\frac{3}{n(n+3)}=\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+2}-\frac{1}{n+3}\right)$$,
which would definitely be a lot more helpful in helping cancel out terms like you're supposed to when doing telescoping series, BUT I don't know why she's doing this. I thought we were supposed to plug in values from n and that's what should be increasing each time, but instead the number being added to n is the one going up and I have no clue why. I don't think I'm asking this question in the best way possible, but I'm kinda confusing myself because she did other examples and they feel nothing like this and I'm just starting to learn all this, so can somebody please give me some insight as to what is going on?
(and I know I'm supposed to also deal with the sum of the $$\frac{1}{2^n}$$ term but I'm kinda ignoring it for now since I don't even know what's going on with the first one
 A: Hint: We can temporarily forget about the $\frac{1}{2^n}$, they sum to $1$.
The terms corresponding to $n=1$, $4$, $7$, $10$, and so on sum to
$$\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}+\frac{1}{10}+\cdots.$$
Telescoping gives sum $1$.
Similarly, the terms corresponding to $n=2,5,8,11,\dots$ sum to $\frac{1}{2}$ and the terms corresponding to $n=3,6,9,12,\dots$ sum to $\frac{1}{3}$.
Now one should redo the argument, by considering partial sums, without change of order. This is because one cannot assume that the radical rearrangement we have done leaves the limit of partial sums unchanged. 
A: What she’s doing with the telescoping part is nice but unnecessary. Without it you can still argue as follows. You’ve rewritten the series like this:
$$\sum_{n\ge 1}\frac3{n(n+3)}=\sum_{n\ge 1}\left(\frac1n-\frac1{n+3}\right)\;.$$
That means that the $m$-th partial sum $s_m$ is
$$s_m=\sum_{n=1}^m\left(\frac1n-\frac1{n+3}\right)\;.$$
This is a finite sum, so it can be rearranged:
$$s_m=\sum_{n=1}^m\left(\frac1n-\frac1{n+3}\right)=\sum_{n=1}^m\frac1n-\sum_{n=1}^m\frac1{n+3}\;.$$
Now notice that as $n$ runs from $1$ through $m$, $n+3$ runs through the consecutive integers from $4$ through $m+4$; thus,
$$\sum_{n=1}^m\frac1{n+3}=\sum_{n=4}^{m+3}\frac1n\;,$$
and
$$s_m=\sum_{n=1}^m\frac1n-\sum_{n=4}^{m+3}\frac1n\;.\tag{1}$$
Every term in the first summation of $(1)$ appears in the second summation except the first three, the terms for $n=1$, $n=2$, and $n=3$. Similarly, every term in the second summation appears in the first except the last three, the terms for $n=m+1$, $n=m+2$, and $n=m+3$. All of those common terms cancel out, and we’re left with
$$\begin{align*}
s_m&=\frac11+\frac12+\frac13-\left(\frac1{m+1}+\frac1{m+2}+\frac1{m+3}\right)\\
&=\frac{11}6-\left(\frac1{m+1}+\frac1{m+2}+\frac1{m+3}\right)\;.
\end{align*}$$
Thus,
$$\begin{align*}
\lim_{m\to\infty}s_m&=\lim_{m\to\infty}\left(\frac{11}6-\left(\frac1{m+1}+\frac1{m+2}+\frac1{m+3}\right)\right)\\\\
&=\frac{11}6-0\\\\
&=\frac{11}6\;,
\end{align*}$$
meaning that $$\sum_{n\ge 1}\frac3{n(n+3)}=\frac{11}6\;.$$
It may be a little easier to see what’s going on if you first do the expansion 
$$\frac1n-\frac1{n+3}=\frac1n-\frac1{n+1}+\frac1{n+1}-\frac1{n+2}+\frac1{n+2}-\frac1{n+3}\;,$$
but it’s not necessary.
The original series $\sum_{n\ge 1}\left(\frac3{n(n+3)}+\frac1{2^n}\right)$ is absolutely convergent, so you can split it as
$$\sum_{n\ge 1}\left(\frac3{n(n+3)}+\frac1{2^n}\right)=\sum_{n\ge 1}\frac3{n(n+3)}+\sum_{n\ge 1}\frac1{2^n}$$
and sum the two simpler series individually. We’ve just dealt with the first one, and the second is geometric, a type for which you have a summation formula.
