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Let \(\phi \) denote the golden ratio, \(\phi = \dfrac{1+\sqrt5}2 \), prove that \(\displaystyle \sum_{n=1}^\infty \dfrac1{\phi^n} = \phi \).

Note by Hummus A
9 months, 3 weeks ago

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Whats \(\phi\)?? -Golden ratio ??
If that's the case then above summation is just the sum of a infinite GP with first term and common ratio \(\frac{1}{\phi}\). \[\therefore \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ \phi ^{ n } } }=\dfrac{\frac{1}{\phi}}{1-\frac{1}{\phi}}=\dfrac{1}{\phi-1}=\phi\] [Since \(\phi^2-\phi-1=0\Rightarrow \phi-1=\dfrac{1}{\phi}\)] Rishabh Cool · 9 months, 3 weeks ago

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@Rishabh Cool Going further, if we define \(x_{k}\) for integers \(k \ge 0\) as the solution to the equation \(\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{x^{n}} = x^{k}\) then \(\displaystyle\lim_{k \to \infty} x_{k} = 1\). Brian Charlesworth · 9 months, 3 weeks ago

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@Rishabh Cool yes,it's the golden ratio Hummus A · 9 months, 3 weeks ago

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That is pretty cool, and \(\phi\) is the unique solution to the equation \(\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{x^{n}} = x\) where \(x \gt 1\).

Writing the sum as \(S = \displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{x}\right)^{n} = \sum_{n=1}^{\infty} y^{n}\) where \(y = \dfrac{1}{x}\) is such that \(0 \lt y \lt 1\), we see that we have an infinite geometric series with

\(S = \dfrac{y}{1 - y} = \dfrac{\dfrac{1}{x}}{1 - \dfrac{1}{x}} = \dfrac{1}{x - 1}\).

Then to have \(S = x\) we require that \(\dfrac{1}{x - 1} = x \Longrightarrow 1 = x^{2} - x \Longrightarrow x^{2} - x - 1 = 0\),

the positive root of which is \(\dfrac{1 + \sqrt{5}}{2} = \phi \gt 1\). Brian Charlesworth · 9 months, 3 weeks ago

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