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

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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

I know not. But it is so interesting!

Log in to reply

i wanted to share the beauty with the brilliant community :)

Log in to reply