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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

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2 Answers 2

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

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  • $\begingroup$ 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 ? $\endgroup$
    – Mainviel
    Commented Nov 13, 2014 at 15:54
  • $\begingroup$ Do you agree that $$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}$$ is a sum involving only the odd integers? $\endgroup$
    – graydad
    Commented Nov 13, 2014 at 15:58
  • $\begingroup$ ah sure, now I see it! thanks $\endgroup$
    – Mainviel
    Commented Nov 13, 2014 at 16:07
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    $\begingroup$ 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. $$\sum_{n=1}^\infty\left(\frac{1}{f_nf_{n+1}}-\frac{1}{f_{n+1}f_{n+2}}\right) = \sum_{n=1}^\infty\frac{1}{f_nf_{n+1}}-\sum_{n=1}^\infty\frac{1}{f_{n+1}f_{n+2}} \\ = 1+\sum_{n=1}^\infty\frac{1}{f_{n+1}f_{n+2}}-\sum_{n=1}^\infty\frac{1}{f_{n+1}f_{n+2}}$$ $\endgroup$
    – graydad
    Commented Nov 14, 2014 at 14:24
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    $\begingroup$ ah I see just because the first two elements are $1$ in the fibonacci sequence! thank you that was very enlightening! $\endgroup$
    – Mainviel
    Commented Nov 14, 2014 at 14:27
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$(a)$. 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.

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  • $\begingroup$ for (a) who do you mean with parity of the denominator? $\endgroup$
    – Mainviel
    Commented Nov 13, 2014 at 15:30
  • $\begingroup$ Depending on whether the denominator $n^s$ is either even or odd. $\endgroup$
    – Lucian
    Commented Nov 13, 2014 at 15:38

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