On sums and identities I am given the following problem set:

(a) the Riemann $\zeta$-function for $s > 1$ is defined through the convergent sum: $$\zeta(s) := \sum_{n = 1}^{\infty} \frac{1}{n^s}$$
  show the identity $$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{3}{4}\zeta(2)$$
(b) show that $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}= \frac{1}{4}$$
(c) we denote $f_n$ as the $n^{th}$ fibonacci term. Show that $$\sum_{n=1}^{\infty} \frac{1}{f_n f_{n+2}} = 1$$

I basically need help on every of those identities since my knowledge about sums is pretty basic. thank you for your hints and help

In the first identity you are interested in the infinite sum of just the odd integers, where $s = 2$. One way you can get this is to observe that $$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \sum_{n=1}^{\infty} \frac{1}{n^2}-\sum_{n=1}^{\infty} \frac{1}{(2n)^2} \\ =\zeta(2)-\sum_{n=1}^{\infty} \frac{1}{4n^2} \\ = \zeta(2)-\frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2} \\ = \zeta(2)-\frac{1}{4}\zeta(2)$$ As for part $(b)$, manipulate the series $$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} = \sum_{n=1}^{\infty} \frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)} \\ = \sum_{n=1}^{\infty} \frac{1}{2n}- \sum_{n=1}^{\infty}\frac{1}{n+1}+ \sum_{n=1}^{\infty}\frac{1}{2(n+2)} \\ =\left(\frac{1}{2}+\frac{1}{4}+\sum_{n=3}^{\infty} \frac{1}{2n}\right)- \left(\frac{1}{2}+\sum_{n=3}^{\infty}\frac{1}{n}\right)+ \left(\sum_{n=3}^{\infty}\frac{1}{2n}\right) $$ For part $(c)$ remember that $f_{n+2} = f_{n+1}+f_n$
