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Golden ratio is everywhere!

\[ \large \sum_{n=1}^\infty \dfrac{(-1)^{n-1}} {n^2 \; \dbinom{2n}{n}} = 2 ( \ln \phi)^2 \]

Prove the equation above.

Notations:


This is a part of the set Formidable Series and Integrals.

Note by Hummus A
7 months ago

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\[\sum_{n=1}^\infty \dfrac{(-1)^{n-1}} {n^2 \binom{2n}{n}} =\sum_{n=1}^\infty \dfrac{(-1)^{n-1}(n-1)!n!} {n*2n!} \\ =\sum_{n=1}^\infty \dfrac{(-1)^{n-1}B(n+1,n)} {n}=\int_0^1 \sum_{n=1}^\infty \dfrac{(-1)^{n-1}x^n(1-x)^{n-1}}{n}dx \\=\int_0^1 \dfrac{\ln(1+x-x^2)}{1-x}dx=\int_0^1 \dfrac{\ln(1+x-x^2)}{x}dx\] The integral is this

Try a similar problem Aareyan Manzoor · 6 months ago

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I know not. But it is so interesting! Joel Yip · 7 months ago

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@Joel Yip i wanted to share the beauty with the brilliant community :) Hummus A · 7 months ago

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