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

Prove the equation above.

**Notations**:

\(\phi \) denote the Golden ratio, \(\phi = \dfrac{1+\sqrt5}2 \approx 1.618 \).

\(\dbinom mn \) denotes the binomial coefficient.

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

## Comments

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

Try a similar problem – Aareyan Manzoor · 1 year, 5 months ago

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I know not. But it is so interesting! – Joel Yip · 1 year, 6 months ago

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– Hummus A · 1 year, 6 months ago

i wanted to share the beauty with the brilliant community :)Log in to reply