How to solve $\sum_{k=1}^\infty \frac{1}{(2k-1)(2k+1)}$

Question

Can you help me please with this sum?

$$\sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)}$$

I have no idea how to solve it. Result is $\frac{1}{2}$.

Thank you

Answer 1

So for nth partial sum we have:
$s_n := \sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)} = \frac12 \sum_{k=1}^{n} \left(\frac1{2k-1}-\frac1{2k+1}\right) = \frac12 \cdot ( 1 - \frac{1}{2n+1})$
Solving the limit:
$ \lim_{n \to \infty } s_n = \lim_{n \to \infty } ( \frac12 \cdot ( 1 - \frac{1}{2n+1}) ) = \frac12 = \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)}$

Thanks for the hint :)

Answer 2

Hint: $$\frac1{(2k-1)(2k+1)}=\frac12\left(\frac1{2k-1}-\frac1{2k+1}\right)$$

Answer 3

Hint
Expand the numerator as such $$1 = \frac12(2k+1 - (2k - 1))$$ And use telescoping series.