How to calculate $\sum_{n=1}^\infty \frac{2}{(2n-1)(2n+1)}$ How can I calculate the sum of the following infinite series: $$\sum_{n=1}^\infty \frac{2}{(2n-1)(2n+1)}$$
 A: Hint : $$\sum_{n=1}^\infty \frac{2}{(2n-1)(2n+1)} = \sum_{n=1}^\infty \left (\frac{1}{2n-1}-\frac{1}{2n+1} \right)$$
A: Since $$\frac{2}{(2n-1)(2n+1)}=\frac{1}{2n-1}-\frac{1}{2n+1}=\int_{2n-1}^{2n+1}\frac{dx}{x^2}$$
your series equals:
$$ \sum_{n\geq 1}\int_{2n-1}^{2n+1}\frac{dx}{x^2}=\int_{1}^{+\infty}\frac{dx}{x^2}=\color{red}{1}.$$
A: Hint:
First thing to do is to try a few terms
$$\frac2{1\cdot3}+\frac2{3\cdot5}+\frac2{5\cdot7}+\frac2{7\cdot9}+\cdots.$$
The partial sums are
$$\frac23,\frac45,\frac67,\frac89\cdots$$
Can't you see a pattern ?

Apparently, the sum of the $n$ first terms is $\dfrac{2n}{2n+1}$. We can check this hypothesis by removing the last term:
$$\frac{2n}{2n+1}-\frac{2}{(2n-1)(2n+1)}=\frac{4n^2-2n-2}{(2n-1)(2n+1)}=\frac{2n-2}{2n-1}=\frac{2(n-1)}{2(n-1)+1}.$$
A: $$ \sum_{i=1}^\infty \frac{2}{(2n-1)(2n+1)}= \sum_{i=1}^\infty \left (\frac{1}{2n-1}-\frac{1}{2n+1} \right)=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+....=1$$
A: This is how you do it formally:
$$\sum_{i=1}^N\frac{2}{(2n-1)(2n+1)} = \sum_{i=1}^N \left (\frac{1}{2n-1}-\frac{1}{2n+1} \right)= \sum_{i=1}^N \left (\frac{1}{2n-1} \right) - \sum_{i=1}^N \left (\frac{1}{2n+1} \right) = 1 + \sum_{i=2}^{N} \left (\frac{1}{2n-1} \right) - \sum_{i=1}^{N-1}\left (\frac{1}{2n+1} \right)  - \frac{1}{2N+1} =1 + \sum_{i=1}^{N-1} \left (\frac{1}{2n+1} \right) - \sum_{i=1}^{N-1} \left (\frac{1}{2n+1} \right)  - \frac{1}{2N+1} =1 - \frac{1}{2N+1} $$
Now take the limit $$\lim_{N\to\infty} \left(1 - \frac{1}{2N+1}\right) =1$$ 
So the series converges towards 1.
