# $\sum\limits_{n\geq1}\frac{x^n}{n{2n\choose n}}=2\sqrt{\frac{x}{4-x}}\arctan\sqrt{\frac{x}{4-x}}$

Context: this is a purely recreational problem that I came up with myself.

I am trying to extend this colsed form:

$$\sum_{n\geq1}\frac{x^n}{n{2n\choose n}}=2\sqrt{\frac{x}{4-x}}\arctan\sqrt{\frac{x}{4-x}},\qquad 0 I know that $$S(x)=\sum_{n\geq1}\frac{x^n}{n{2n\choose n}}$$ converges for $$|x|<4$$, but the closed form above only works for half that range. Is there a closed form for the other half?

I think that it might come down to figuring out how to integrate $$\int_0^1\frac{\mathrm dt}{xt^2-xt+1}$$ when $$-4, but I'm not sure. In your answer, please show your proof. Thanks.

For those interested, a proof.

Noting that $${2n\choose n}^{-1}=\frac{n\Gamma^2(n)}{2\Gamma(2n)}$$, $$S(x)=\frac12\sum_{n\geq1}\int_0^1x^n[t(1-t)]^{n-1}\mathrm dt$$ Then using the geometric series, $$S(x)=\frac{x}2\int_0^1\frac{\mathrm dt}{xt^2-xt+1}$$ Then we focus on $$I(k;a,b,c)=\int_0^k\frac{\mathrm dt}{at^2+bt+c}$$ If we complete the square in the denominator then preform the appropriate trig substitution, we see that for $$4ac>b^2$$, $$I(k;a,b,c)=\frac2{\sqrt{4ac-b^2}}\bigg[\arctan\frac{2ak+b}{\sqrt{4ac-b^2}}-\arctan\frac{b}{\sqrt{4ac-b^2}}\bigg]$$ So for $$0, $$S(x)=\frac{x}2I(1;x,-x,1)$$ $$S(x)=2\sqrt{\frac{x}{4-x}}\arctan\sqrt{\frac{x}{4-x}}$$ As desired.

• $-i\arctan(ix)=\text{arctanh}(x)=\frac{1}{2}\log\frac{1+x}{1-x}$ allows to cover the whole $(-4,4)$. – Jack D'Aurizio Jan 11 at 17:25
• You are actually just integrating the Maclaurin series of the squared arcsine. – Jack D'Aurizio Jan 11 at 17:26
• The integral you asked for equals $\frac{4 \tanh ^{-1}\left(\sqrt{\frac{x}{x-4}}\right)}{\sqrt{(x-4) x}}$ – ablmf Jan 11 at 21:23
• @ablmf how do you find this? – clathratus Jan 11 at 21:46
• @clathratus Mathematica. BTW, the anti-derivative is $\frac{2 \tan ^{-1}\left(\frac{(2 t-1) \sqrt{x}}{\sqrt{4-x}}\right)}{\sqrt{-(x-4) x}}$. So this is the proof. – ablmf Jan 11 at 21:51