<s>[INST] How to compute fraction sums? For example, $$\sum\limits_{k=1}^{n}\frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}$$
Is there an easier way to evaluate fraction sums (without using partial sums)?
 [/INST] A general approach which works quite well is partial fraction decomposition:
$$
\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\left(\frac{1}{(2k-1)}-\frac{1}{(2k+1)}\right)
$$
After this, you can use the technique of telescoping.
In your example, this yields:
\begin{align}
& \sum_{k=1}^{n}{\frac{1}{(2k-1)(2k+1)}} = \sum_{k=1}^{n}{\frac{1}{2}\left(\frac{1}{(2k-1)}-\frac{1}{(2k+1)}\right)} \\
&= \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+…+\frac{1}{2n-3}-\frac{1}{2n-1}+\frac{1}{2n-1}-\frac{1}{2n+1}\right) \\
&= \frac{1}{2}\left(1-\frac{1}{2n+1}\right) = \frac{n}{2n+1}
\end{align}
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