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The central binomial $${2n\choose n}=\frac{(2n)!}{(n!)^2}$$

The series for natural logarithmic of $\ln(2)$ is

$$\ln(2)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$$

and the integral for it is $$\int_0^1\frac{1}{1+x}dx=\ln(2)$$

Show that,

$$\sum_{n=1}^{\infty}\frac{\left(-\frac{1}{2}\right)^{n-1}}{{n^2}{2n\choose n}}=\ln^2(2)$$

Can anybody help me to prove this sum, I found it through mathematical experimental.

I guess if there is an integral for, $\frac{1}{n^2{2n\choose n}}=\int_0^1f(x,n)dx$ it might be helpful.

I know that $\frac{2^n}{(2n+1){2n\choose n}}=\int_0^1[2x(1-x)]^ndx$

  • A link for some inspiration (you are encouraged too to provide answers). Excellent continuation (with pure imaginary!), – Raymond Manzoni May 09 '16 at 08:09
  • See also http://math.stackexchange.com/questions/99809/how-to-prove-by-arithmetical-means-that-sum-limits-k-1-infty-frack-1/128680#128680 – Marco Cantarini May 09 '16 at 12:04
  • @MarcoCantarini (+1), I read it still not sure how to use it to solve this problem, any more help –  May 10 '16 at 04:32
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    @pisquare In the links posted by me and Raymond Manzoni there are proofs of the identity $\frac{1}{2}\sum_{k\ge1}\frac{1}{k^{2}\binom{2k}{k}}(2x)^{2k}=\arcsin^{2}(x).$ So you have only to take $x=\frac{i}{2\sqrt{2}}$ and use the definition $(1)$ of http://mathworld.wolfram.com/InverseSine.html – Marco Cantarini May 10 '16 at 08:28
  • @Marco Cantarini (+1), I see! –  May 10 '16 at 10:52
  • Can you explain what "I found it through mathematical experimental" really means? – Jack D'Aurizio May 11 '16 at 11:34
  • Simple Mr jack. 2,4,6,8,...2n –  May 11 '16 at 13:11
  • Hi @JackD'Aurizio also this account too. I want to end this maths staff here, wasting a lot of my in writing and not doing my quest for finding new formulas. –  May 11 '16 at 13:16
  • I don't get it. How can you claim a specific value for a series by checking (what?) for $2,4,6,8,\ldots,2n$? – Jack D'Aurizio May 11 '16 at 13:42
  • Mr Jack I am starting to see that you are nice man. I am apologising for insulting you. I can see your proved of my proposed questions is wonderful concise and full of detail. Explaining how i got these formulas from maths experimental and guessing is quite not easy to explain. I may find an example in the future and show it to you. –  May 11 '16 at 14:04

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Using these formulas: $$(\arcsin(x))^2=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$$

$$\arcsin(x)=i\ln(ix+\sqrt{1-x^2})$$

$$-2\ln^2(ix+\sqrt{1-x^2})=\sum_{n=0}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$$

Let $x=\frac{i}{2\sqrt2}$

Simplify it and you should get $$\sum_{n=1}^{\infty}\frac{\left(-\frac{1}{2}\right)^{n-1}}{{n^2}{2n\choose n}}=\ln^2(2)$$