Some sum

Either by using Newton's generalised binomial theorem, or by differentiating the generating function for Catalan numbers, it can be shown that $\displaystyle \frac{1}{\sqrt{1-4u}} = \sum_{n=0}^\infty \binom{2n}{n}u^n$ .

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Method 1 :

Note that $\displaystyle \sum_{n=0}^\infty \binom{4n}{2n}x^{2n} = \frac{1}{2} \left[ \sum_{n=0}^\infty \binom{2n}{n}x^n + \sum_{n=0}^\infty \binom{2n}{n}(-x)^n \right] = \frac{1}{2} \left[ \frac{1}{\sqrt{1-4x}} + \frac{1}{\sqrt{1+4x}} \right]$ . Since in the given sum, $x = \dfrac{1}{5}$ , this gives us $\displaystyle \frac{1}{2} \left[ \frac{1}{\sqrt{1-4/5}} + \frac{1}{\sqrt{1+4/5}} \right] = \boxed{\sqrt{\dfrac{20}{9}}}$ .

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Method 2 :

To isolate the even terms, we can substitute $u = ix$ and take the real part to get

$\sum_{n=0}^\infty \binom{4n}{2n}(-x^2)^n = \mathrm{Re} \left( \sum_{n=0}^\infty \binom{2n}{n}(ix)^n \right) = \mathrm{Re} \left( \frac{1}{\sqrt{1-4ix}} \right) .$

If we let $\dfrac{1}{\sqrt{1-4ix}} = re^{i\theta}$ , then we have $\mathrm{Re} \left( \dfrac{1}{\sqrt{1-4ix}} \right) = r\cos \theta$ . $1-4ix = \dfrac{e^{-2i\theta}}{r^2}$ , so we have $\cos 2\theta = r^2$ and $\sin 2\theta = 4r^2x$ . Hence, $\theta = \dfrac{1}{2} \arctan 4x$ and $r = \dfrac{1}{\sqrt[4]{1+16x^2}}$ .

Thus,

$\begin{aligned} \mathrm{Re} \left( \frac{1}{\sqrt{1-4ix}} \right) &= r\cos \theta \\ &= \frac{\cos(\frac{1}{2} \arctan 4x)}{\sqrt[4]{1+16x^2}} \\ &= \frac{\sqrt{1+\cos \arctan 4x}}{\sqrt{2} \sqrt[4]{1+16x^2}} \\ &= \frac{\sqrt{1+\sqrt{\dfrac{1}{1+16x^2}}}}{\sqrt{2} \sqrt[4]{1+16x^2}} \\ &= \sqrt{\frac{1+\sqrt{1+16x^2}}{2(1+16x^2)}} . \end{aligned}$

By replacing $-x^2$ with $x$ , we therefore have the result $\displaystyle \sum_{n=0}^\infty \binom{4n}{2n}x^n = \sqrt{\frac{1+\sqrt{1-16x}}{2(1-16x)}}$ . Since in our given sum $x = \dfrac{1}{25}$ , as a result we have

$\sum_{n=0}^\infty \frac{\binom{4n}{2n}}{25^n} = \sqrt{\frac{1+\sqrt{1-16/25}}{2(1-16/25)}} = \boxed{\sqrt{\dfrac{20}{9}}} .$