<s>[INST] Finding sum $\sum_{i=1}^n \frac1{4i^2-1}$ I have been having problem with calculating the following summation:
$$
\sum_{i=1}^n  {1\over 4i^2-1} = {1\over3} + {1\over15} + {1\over35} + \cdots + {1\over 4n^2-1}
$$
I do know the answer, but just can not find the way to get it.
Thank you in advance.
 [/INST] You have, 
$$
\sum_{i=1}^n  {1\over 4i^2-1} = \sum_{i=1}^n\frac{1}{{(2i + 1)(2i - 1)}} 
$$
$$
= \sum_{i=1}^n \left(\frac{1}{{2(2i - 1)}} - \frac{1}{{2(2i + 1)}}\right)
= \frac{1}{2}\left(\sum_{1\leq i\leq n} \frac{1}{{(2i - 1)}} - \sum_{1\leq i\leq n}\frac{1}{{(2i + 1)}}\right)
$$
$$
=\frac{1}{2}\left( \frac{1}{1}+\sum_{2\leq i\leq n} \frac{1}{{(2i - 1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right).
$$
By translation of index i=k+1, you have,
$$
=\frac{1}{2}\left( \frac{1}{1}+\sum_{2\leq k+1\leq n} \frac{1}{{(2[k+1] - 1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right).
$$
Remember that $2\leq k+1\leq n$ if, only if, $2-1\leq k\leq n-1$. Then
$$
=\frac{1}{2}\left( \frac{1}{1}+\sum_{1\leq k\leq n-1} \frac{1}{{(2k+1)}} - \sum_{1\leq i\leq n-1}\frac{1}{{(2i + 1)}}- \frac{1}{2n+1}\right)
$$
$$
=\frac{1}{2}-\frac{1}{4n+2}.
$$
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