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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
11 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 · 10 months ago

I know not. But it is so interesting! · 11 months ago