Finding the closed form of a sum 4 I've been asked to find the closed type of this sum:
$$\sum_{i=1}^{n} \frac{1}{(2i-1)(2i+1)}\quad \;$$
My first thought was to break it in two and find the closed form of the other two sums:
$$\sum_{i=1}^{n} \frac{1}{(2i-1)(2i+1)}\quad=\frac{1}{2}\sum_{i=1}^{n} \frac{1}{2i-1}\quad-\frac{1}{2}\sum_{i=1}^{n} \frac{1}{2i+1}\quad \;$$
The problem is I can't think of a way to find these two closed forms, could you help me find a way? Or is it not possible to solve it by breaking it into two sums?
 A: It's a telescopic sum. Do
$$\sum_{i=1}^n \frac{1}{(2i-1)(2i+1)}=\sum_{i=1}^n \frac{1}{2}\left(\frac{1}{2i-1}-\frac{1}{2i+1}\right)=\frac{1}{2} \left(1-\frac{1}{2n+1}\right)=\frac{n}{2n+1}$$
A: Note: As a comment indicated: Your approach of breaking the sums is also quite ok. Since, the sums are finite there is no question regarding divergence. You could proceed as follows

\begin{align*}
\sum_{i=1}^{n} \frac{1}{(2i-1)(2i+1)}
&=\frac{1}{2}\sum_{i=1}^{n} \frac{1}{2i-1}-\frac{1}{2}\sum_{i=1}^{n} \frac{1}{2i+1}\\
&=\frac{1}{2}\sum_{i=0}^{n-1} \frac{1}{2i+1}-\frac{1}{2}\sum_{i=1}^{n} \frac{1}{2i+1}\tag{1}\\
&=\frac{1}{2}\left(1+\sum_{i=1}^{n-1}\frac{1}{2i+1}\right)-\frac{1}{2}\left(\sum_{i=1}^{n-1} \frac{1}{2i+1}+\frac{1}{2n+1}\right)\tag{2}\\
&=\frac{1}{2}\left(1-\frac{1}{2n+1}\right)\\
&=\frac{n}{2n+1}
\end{align*}

Comment:


*

*In (1) we shift the index $i$ of the first sum by 1 to better see what's going on

*In (2) we split the first summand from the left sum and the last summand from the right sum. We observe the sums in the middle cancel out.
In fact the line tagged with (2) was only written for demonstration and can be skipped without problems. 
