Confluent hypergeometric function

Singular series posted by @infseriesbot

Today we evaluate this series posted by @infseriesbot \[ \left(\sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n\right)^2 = e^{4x}\sum_{n=0}^{\infty} \frac{(2n)!}{n!^4} x^{2n}\] The proof will rely on some properties of the confluent hypergeometric function and the binomial coefficient

Proof:

We start with the left hand side. Recall that the Pocchhammer symbol or rising factorial can be expressed as quotient of two gamma functions \[ (x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}\] Hence \[ (2n)! = \Gamma(2n+1) = (1)_{2n}\] Pochhammer symbols also satisfy the duplication formula: \[(x)_{2n} = 4^n\left(\frac{x}{2}\right)_{n}\left(\frac{1+x}{2}\right)_{n}\] Hence \[ (2n)! = 4^{n}\left(\frac{1}{2}\right)_{n}(1)_{n}\] Note also that $\displaystyle (1)_{n} = n!$. Hence, \[\sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n =\sum_{n=0}^{\infty} \frac{4^{n}\left(\frac{1}{2}\right)_{n}(1)_{n}}{(1)_{n}(1)_{n}} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n}}{(1)_{n}} \frac{(4x)^n}{n!} = {}_{1}F_{1}\left(\frac{1}{2};1;4x\right)\] The right hand side is the confluent hypergometric function ${}_{1}F_{1}(a;b;x)$. This function is also denoted with $M(a;b;x)$:

The confluent hypergometric function satisfies the Kummer's second theorem or second transformation: \[ {}_1F_1(a,2a,x)= e^{x/2}\, {}_0F_1 \left(; a+\tfrac{1}{2}; \tfrac{x^2}{16} \right) \] Therefore \[ \sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n= {}_{1}F_{1}\left(\frac{1}{2};1;4x\right) =e^{2x}\, {}_0F_1 \left(; 1; x^2 \right)\] From the definition of ${}_{0}F_{1}$: \[e^{2x}\, {}_0F_1 \left(; 1; x^2 \right) = e^{2x} \sum_{n=0}^{\infty} \frac{x^{2n}}{n!^2} \] Hence \[ \sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n = e^{2x} \sum_{n=0}^{\infty} \frac{x^{2n}}{n!^2} \] Squaring both sides \[ \left(\sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n\right)^2 = e^{4x} \left(\sum_{n=0}^{\infty} \frac{x^{2n}}{n!^2} \right)^2 \tag{1} \] We can evaluate the right hand side with the Cauchy product: \begin{align*}\left(\sum_{n=0}^{\infty} \frac{x^{2n}}{n!^2} \right)^2 = &\sum_{n=0}^{\infty}\sum_{j=0}^{n} \frac{1}{j!^2(n-j)!^2} x^{2n}\\ =& \sum_{n=0}^{\infty}\frac{1}{n!^2}\sum_{j=0}^{n} \frac{n!^2}{j!^2(n-j)!^2} x^{2n}\\ =& \sum_{n=0}^{\infty}\frac{x^{2n}}{n!^2}\sum_{j=0}^{n} {\binom{n}{j}}^2 \tag{2} \end{align*} From the properties of the binomial coefficient we know that: \[ \sum_{j=0}^{J}{\binom{n}{j}}^2 = \binom{2n}{n} \quad J\geq n \tag{3} \] Therefore \[\left(\sum_{n=0}^{\infty} \frac{x^{2n}}{n!^2} \right)^2 = \sum_{n=0}^{\infty}\frac{x^{2n}}{n!^2}\sum_{j=0}^{n} {\binom{n}{j}}^2 = \sum_{n=0}^{\infty}\frac{x^{2n}}{n!^2}\binom{2n}{n} = \sum_{n=0}^{\infty}\frac{(2n)!}{n!^4}x^{2n} \tag{4}\] Hence from (1),(2),(3) and (4) we can conclude: \[ \boxed{\left(\sum_{n=0}^{\infty} \frac{(2n)!}{n!^3} x^n\right)^2 = e^{4x}\sum_{n=0}^{\infty} \frac{(2n)!}{n!^4} x^{2n}}\]