# Computing the series $\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \Big(\frac{|A|}{2}\Big)^{2n}$

I want to know how to compute the closed form for the series

$$\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \Big(\frac{|A|}{2}\Big)^{2n}, \quad |A| < 1,$$

with or without special functions. When I plugged this into Mathematica, it gave me

$$\frac{1}{\sqrt{1 - |A|^2}}.$$

I also would really, really like to know if there are exercises/lessons all in one place (it's a bit hard to search the SE for something like this) or texts out there that can help me learn methods of doing these computations. I know I won't be able to always find closed forms of expressions like this, but I'm always amazed at what I see folks are able to compute here on Stack Exchange.

I thought to use a modified Bessel function $$I_\alpha(z)$$ with $$\alpha = 0$$,

$$I_0(z) = \sum_{k=0}^\infty \frac{1}{(k!)^2} \left(\frac{z}{2}\right)^{2k}$$

and perhaps differentiate and then evaluate at $$|A|$$, but I haven't figured out how to manipulate that. Any suggestions, general and specific?

Note: For some context, I am generally trying to compute norms for squeezed coherent states (quantum mechanics application) which have the form $$\psi(z) = e^{Az^2/2} e^{-|z|^2/2}$$, $$z \in \mathbb{C}$$, and I wind up with series like this all the time.

• Using the binomial theorem for the square root you should be able to find your series. You can find the formula even on Wikipedia.. Maybe this could help you find a way back from your series to the closed form. Commented Jul 6, 2021 at 19:21
• Ah yes, great suggestion. I had forgotten about the general binomial theorem. I'll try to play with this. Commented Jul 6, 2021 at 20:08
• Are you familiar with generating functions and methods used in finding them? Commented Jul 6, 2021 at 20:34
• @A-LevelStudent I've only read some about generating functions and understand the general idea of them, but I have little to no explicit experience working with them. Commented Jul 6, 2021 at 20:51
• @CyCeez Ok, I only have a small amount of experience with them myself, but what I've encountered is quite straightforward. I'll try to type up an answer for you involving them. Commented Jul 6, 2021 at 20:53

Applying the Taylor development to the function $$f(A)=\frac{1}{\sqrt{1-A^2}}$$ \begin{align} f(A) &= (1-A^2)^{-\frac{1}{2}} \\ &= 1 + \frac{1}{1!} \left(-\frac{1}{2}\right)(-A^2)+ \frac{1}{2!}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)(-A^2)^2 +...+\frac{1}{n!}\left(\prod_{i=0}^{n-1}\left(-\frac{1}{2}-i\right)\right)(-A^2)^n+...\\ &=1 + \sum_{n=1}^{+\infty}\left(\frac{1}{n!}(-A^2)^n\prod_{i=0}^{n-1}\left(-\frac{1}{2}-i\right)\right) \\ &=1 + \sum_{n=1}^{+\infty}\frac{(2n-1)!!}{2^nn!}(A^2)^n \\ &=1 + \sum_{n=1}^{+\infty}\left(\frac{(2n-1)!!}{2^nn!}\times\frac{(2n)!!}{2^nn!}\right)(A^2)^n \\ &=1 + \sum_{n=1}^{+\infty}\frac{(2n)!}{(2^n)^2(n!)^2}(A^2)^n \\ &=\sum_{n=0}^{+\infty}\frac{(2n)!}{(n!)^2}\left(\frac{A^2}{2}\right)^n \\ \end{align}

Q.E.D

Let $$y=\sum_{n=1}^\infty \frac{(2n)!}{(n!)^2} x^{2n}=\sum_{n=1}^\infty\binom{2n}{n}x^{2n}.$$ Before we begin, note that $$\binom{2n+2}{n+1}=\frac{2(2n+1)}{n+1}\binom{2n}{n}.$$ The strategy I will use here is to manipulate the sum in two different ways to make the $$n$$th coefficient equal to $$\binom{2n+2}{n+1}$$ instead of $$\binom{2n}{n}$$. Here goes:

(Note: In order to make this solution as clear as possible I have included every step that is relevant; the solution is actually quite straightforward and if you persevere you will reasonably quickly get to the end :-) )

\begin{align}y&=\sum_{n=1}^\infty \frac{(2n)!}{(n!)^2} x^{2n}\\ \implies xy&=\sum_{n=1}^\infty \frac{(2n)!}{(n!)^2} x^{2n+1}\\ \implies \frac{d}{dx}(xy)=y+x\frac{dy}{dx}&=\sum_{n=1}^\infty (2n+1)\binom{2n}{n}x^{2n}\\ \implies xy+x^2\frac{dy}{dx}&=\sum_{n=1}^\infty (2n+1)\binom{2n}{n}x^{2n+1}\\ \implies\int xy+x^2\frac{dy}{dx}~dx&=C_1+\sum_{n=1}^\infty\frac{(2n+1)}{2(n+1)}\binom{2n}{n}x^{2n+2}=C_1+\frac{1}{4}\sum_{n=1}^{\infty}\binom{2n+2}{n+1}x^{2n+2} .\end{align} However, we can also write $$\sum_{n=1}^\infty \binom{2n+2}{n+1}x^{2n+2}=y-2x^2.$$ Hence, \begin{align}\int xy+x^2\frac{dy}{dx}~dx&=C_1+\frac{1}{4}(y-2x^2)\\ \implies xy+x^2\frac{dy}{dx}&=\frac{1}{4}\frac{dy}{dx}-x\\ \implies \frac{dy}{dx}=(y+1)\cdot\frac{x}{\frac{1}{4}-x^2}.\end{align} Separating the variables: \begin{align}\int\frac{1}{1+y}~dy&=\int\frac{x}{\frac{1}{4}-x^2}dx\\ \implies \ln(1+y)&=C_2-\frac{1}{2}\ln\left(\frac{1}{4}-x^2\right)\\ \implies\ln(1+y)&=\ln\frac{1}{2}+\ln\frac{1}{\sqrt{\frac{1}{4}-x^2}}\\ \implies 1+y&=e^{\ln\frac{1}{2}}\cdot e^{\ln\frac{1}{\sqrt{\frac{1}{4}-x^2}}}\\ &=\frac{1}{\sqrt{1-4x^2}}\\ \implies \binom{2\cdot0}{0}x^{2\cdot0}+\sum_{n=1}^\infty\binom{2n}{n}x^{2n}&=\frac{1}{\sqrt{1-4x^2}}\\ \implies \sum_{n=0}^\infty\binom{2n}{n}x^{2n}&=\frac{1}{\sqrt{1-4x^2}}.\end{align} Plug in $$x=\displaystyle\frac{\lvert A\rvert}{2}$$ and you have your answer.

I hope that helps. If you have any questions please don't hesitate to ask :)

• Wow, that's so great. Thank you. My only question is I'm not sure how you arrived at $\sum_n \binom{2n}{n} = y - x^2$. Commented Jul 7, 2021 at 4:31
• @CyCeez You're welcome, I'm really glad you found it useful. For more examples of using this sort of method you may want to have a look at my question: math.stackexchange.com/questions/4144655/… That was a mistake which I have now corrected, thanks for spotting that! Commented Jul 7, 2021 at 7:57

Just for reference, I guess, I finally did go back and take the suggestion of @samario28 and use the generalized binomial theorem,

$$(1 + z)^\alpha = \sum_{n=0}^\infty \frac{(\alpha)_n}{n!} z^n = \sum_{n=0}^\infty \binom{\alpha}{n} z^n,$$

where $$(\alpha)_n$$ denotes the falling factorial. Just for good measure, a ratio test of $$a_n = (2n)!/(n!)^2 z^{2n}$$ will show that radius of convergence is for $$|z| < 1/2$$, which corresponds to our values for $$|A|$$. We have

\begin{align*} \binom{2n}{n} &= \frac{(2n)!}{n!n!} \\ &= \frac{2n (2n-1)(2n-2) \cdots 3 \cdot 2 \cdot 1}{n!n!} \\ &= \frac{2^{2n} n (n-\tfrac{1}{2}) (n-1) (n- \frac{3}{2}) \cdots \frac{3}{2} \cdot 1 \cdot \frac{1}{2} \cdot 1}{n!n!} \\ &= \frac{2^{2n} (n-\tfrac{1}{2}) (n- \frac{3}{2}) \cdots \frac{3}{2} \cdot \frac{1}{2}}{n!} \\ &= \frac{2^{2n} (-1)^n (\tfrac{1}{2}-n) (\frac{3}{2}-n) \cdots (-\frac{3}{2}) \cdot (-\frac{1}{2})}{n!} \\ &= \frac{2^{2n} (-1)^n (-\frac{1}{2})(-\frac{1}{2}-1) \cdots (-\frac{1}{2}-n+2)(-\frac{1}{2} - n+1)}{n!} \\ &= \frac{2^{2n} (-1)^n (-\frac{1}{2})_{n}}{n!}. \end{align*}

Thus, $$\alpha = -1/2$$, and so

\begin{align*} \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \left(\frac{|A|}{2}\right)^{2n} &= \sum_{n=0}^\infty \frac{\big(-\!\frac{1}{2}\big)_n}{n!} \cdot 2^{2n} (-1)^n \left(\frac{|A|}{2}\right)^{2n} \\ &= \sum_{n=0}^\infty \frac{\big(-\!\frac{1}{2}\big)_n}{n!} \cdot (-|A|^2)^n \\ &= \frac{1}{\sqrt{1 - |A|^2}}. \end{align*}