Convergence of power series with factorial ratio coefficients

Question

$$\sum _{n=0}^{\infty }\:2^{2n}\cdot \frac{\left(n!\right)^2}{\left(2n\right)!}x^n$$

Using ratio test we can see that radius of convergence is $R = 1$. Though I'm not sure how to find the exact function in a closed form, like $\frac{1}{1-x}$ with $\sum _{n=0}^{\infty }\:x^n$. My idea is to find some relation with other functions, like $f(x) = (1-x^2)f'(x)$ and then deduce the answer, but I'm kind of stuck. Could you give hints for this problem?

Answer 1

The ratio test gives:$$\lim_{\text{n}\to\infty}\left|\frac{2^{2\left(\text{n}+1\right)}\cdot\frac{\left(\text{n}+1\right)!}{\left(2\left(\text{n}+1\right)\right)!}\cdot x^{\left(\text{n}+1\right)}}{2^{2\text{n}}\cdot\frac{\text{n}!}{\left(2\text{n}\right)!}\cdot x^\text{n}}\right|=\lim_{\text{n}\to\infty}\left|\frac{2x}{1+2\text{n}}\right|=0\tag1$$

And the closed-form of your sum is dependent on the error-function.

Answer 2

You can investigate the function that your series represents as follows:

Let $$f(x) = \sum_{n\geqslant 0} \frac{(2^n n!)^2}{(2n)!}x^n $$ and define the function $$\begin{align}g(t) = \frac{f(t^2)-1}{t}, \tag 1 \end{align}$$ with $ g(0)=0 $. The power series for $g$ is easily deduced, $$ g(t)=\sum_{n \geqslant 1} \frac{(2^nn!)^2}{(2n)!} t^{2n-1}. $$ As above, this series converges for $|t| \leqslant 1 $ and in this interval, $$\begin{align}tg'(t) &= \sum_{n\geqslant 1} (2n-1)\frac{(2^nn!)^2}{(2n)!}t^{2n-1} \\ &=2t +\sum_{n\geqslant2}(2n) \frac{(2^{n-1}(n-1)!)^2}{(2n-2)!}t^{2n-1}\\ &=\frac{d}{dt}\left(t^2+t^3\sum_{n\geqslant 2}\frac{(2^{n-1}(n-1)!)^2}{(2n-2)!}t^{2n-3} \right) \\ &= \frac{d}{dt}\left( t^2+t^3 \sum_{n\geqslant 1} \frac{(2^nn!)^2}{(2n)!}t^{2n-1} \right) \\ &=\frac{d}{dt} \left( t^2+t^3 g(t) \right) \end{align}.$$

This yields a first order differential equation for $g(t)$, which can be arranged as, $$(1-t^2)g'(t)-3t g(t) = 2.$$ If we now apply an integrating factor of $(1-t^2)^{1/2}$, we obtain, $$\frac{d}{dt} \left( (1-t^2)^{3/2}g(t) \right) = 2(1-t^2)^{1/2} = \frac{d}{dt}\left( \sin^{-1} (t) + t(1-t^2)^{1/2} \right). $$ This gives, using $g(0) = 0$, $$ g(t) = \frac{\sin^{-1}(t)+t(1-t^2)^{1/2}}{(1-t^2)^{3/2}} $$ Inverting the definition of $g(t) $ gives, for $x \geq 0$, $$ f(x) = 1+\sqrt{x} \cdot \frac{ \sin^{-1}( \sqrt x )+ \sqrt{x(1-x)}}{(1-x)^{3/2}}. $$ Last, a similar argument is needed to derive the closed form for $f(x)$ when $x \leq 0$, by adapting equation $(1)$. I believe the result is $$ f(x) = 1-\sqrt{ |x| }\cdot \frac{\log(\sqrt{|x|} + \sqrt{1+|x|})+\sqrt{|x|(1+|x|)}}{(1+|x|)^{3/2}}. $$ Combined, we find, $$ f(x) = 1 + \frac{ x \sqrt{1-x} + \left\{ {\sqrt{x} \sin^{-1}\sqrt{x}, \quad\quad\quad\quad x \geqslant 0 \atop \sqrt{-x} \log(\sqrt{1-x}-\sqrt{-x}), \quad x \leqslant 0} \right . }{(1-x)^{3/2}} $$ This is fairly complicated and I hope contains no mistakes. Please let me know if I am wrong!

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Additional comment (1): I have checked numerically the closed form solution against the original power series over $-\frac{1}{2} \leqslant x \leqslant \frac{1}{2}$ with agreement to 5 decimal places, so now reasonably confident the formula is OK.

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Additional comment (2): with acknowledgement to the comment below, when $x\leqslant 0$ the first expression continues to apply if the root of $\sqrt{x}$ is allowed complex values. In this case, writing $\sqrt{x} = i\theta$, we obtain, $$ (i\theta) \sin^{-1} i\theta = \theta \log (\sqrt{1+\theta^2} -\sqrt{\theta}) $$