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# Wow

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 $$\phi^2-\phi-1=0\Rightarrow \phi-1=\dfrac{1}{\phi}$$] · 9 months, 3 weeks ago

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 $$\displaystyle\lim_{k \to \infty} x_{k} = 1$$. · 9 months, 3 weeks ago

yes,it's the golden ratio · 9 months, 3 weeks ago

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 $$\dfrac{1 + \sqrt{5}}{2} = \phi \gt 1$$. · 9 months, 3 weeks ago