# 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 $f_{n+2} = f_{n+1}+f_n$

• I have trouble understanding the first step $$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \sum_{n=1}^{\infty} \frac{1}{n^2}-\sum_{n=1}^{\infty} \frac{1}{(2n)^2}$$ how do we know/get that equality ? – Mainviel Nov 13 '14 at 15:54
• Do you agree that $$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}$$ is a sum involving only the odd integers? – graydad Nov 13 '14 at 15:58
• ah sure, now I see it! thanks – Mainviel Nov 13 '14 at 16:07
• So you rewrote it as another telescoping sum. That is very useful. Just manipulate the sum into a form where it's clear that everything will cancel out. $$\sum_{n=1}^\infty\left(\frac{1}{f_nf_{n+1}}-\frac{1}{f_{n+1}f_{n+2}}\right) = \sum_{n=1}^\infty\frac{1}{f_nf_{n+1}}-\sum_{n=1}^\infty\frac{1}{f_{n+1}f_{n+2}} \\ = 1+\sum_{n=1}^\infty\frac{1}{f_{n+1}f_{n+2}}-\sum_{n=1}^\infty\frac{1}{f_{n+1}f_{n+2}}$$ – graydad Nov 14 '14 at 14:24
• ah I see just because the first two elements are $1$ in the fibonacci sequence! thank you that was very enlightening! – Mainviel Nov 14 '14 at 14:27

$(a)$. Split the infinite series into two subseries, depending on the parity of each term's denominator.

$(b)$. It's a telescoping series.

$(c)$. Use the general formula for the $n^{th}$ Fibonacci number.

• for (a) who do you mean with parity of the denominator? – Mainviel Nov 13 '14 at 15:30
• Depending on whether the denominator $n^s$ is either even or odd. – Lucian Nov 13 '14 at 15:38