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Golden ratio:$\phi$ $$5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}=\zeta(2)\tag1$$

We'll known Basel's problem

$$\sum_{k=1}^{\infty}{1\over k^2}=\zeta(2)\tag2$$

How do I transform $(1)$ to $(2)?$

Extra information:

$$\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}=1\tag3$$

Nothing come to my mind! Any hints?

  • 0
    I'd say that if $0 < a < b < 1$ then $\sum_{k=1}^\infty \frac{b^{k}-a^{k}}{k^{2}}= \sum_{k=1}^\infty \frac{1}{k^{2}}\int_a^b k x^{k-1}dx = \int_a^b \sum_{k=1}^\infty \frac{ x^{k-1}}{k}dx=\int_a^b \frac{-\ln(1-x)}{x}dx = ...$ – 2017-01-21

1 Answers 1

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You want to show that

$5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2} =\zeta(2)\tag1 $.

The function you need is the dilogarithm, defined by

$L_2(z) =\sum_{k=1}^{\infty}\dfrac{z^k}{k^2} $.

There is much useful info here:

https://en.wikipedia.org/wiki/Spence's_function

So you want to show that

$5(L_2(\phi^{-1})-L_2(\phi^{-2})) =\zeta(2) $.

Since $\phi =\dfrac{1+\sqrt{5}}{2} $, $\phi-1 =\dfrac{\sqrt{5}-1}{2} $, $1-\phi =\dfrac{1-\sqrt{5}}{2} =\dfrac{-1}{\phi} $ and $\dfrac{1}{\phi^2} =\dfrac{(1-\sqrt{5})^2}{4} =\dfrac{1-2\sqrt{5}+5}{4} =\dfrac{3-\sqrt{5}}{2} $, this is equivalent to $5(L_2(\dfrac{\sqrt{5}-1}{2})-L_2(\dfrac{3-\sqrt{5}}{2})) =\zeta(2) $.

According to the Wikipedia page above, $L_2(\dfrac{\sqrt{5}-1}{2}) =\dfrac{\pi^2}{10}-\ln^2(\dfrac{\sqrt{5}-1}{2}) $ and $L_2(\dfrac{3-\sqrt{5}}{2})) =\dfrac{\pi^2}{15}-\ln^2(\dfrac{\sqrt{5}-1}{2}) $.

Subtracting these, the $\ln^2$ terms cancel out and we get

$\begin{array}\\ L_2(\dfrac{\sqrt{5}-1}{2})-L_2(\dfrac{3-\sqrt{5}}{2})) &=\dfrac{\pi^2}{10}-\dfrac{\pi^2}{15}\\ &=\dfrac{\pi^2}{30}\\ \text{so}\\ 5(L_2(\dfrac{\sqrt{5}-1}{2})-L_2(\dfrac{3-\sqrt{5}}{2}))) &=\dfrac{\pi^2}{6}\\ &=\zeta(2)\\ \end{array} $