Accurate identities related to $\sum\limits_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}x^n$ and $\sum\limits_{n=0}^{\infty}\frac{(2n)!}{(n!)^4}x^n$

Solution 1:

Herein we present solutions to the $(1)$ and $(3)$ in the OP based on exploiting the uniform convergence of the series for $e^x$ on compact intervals. To that end, we proceed.

To show that $(1)$ in the OP is correct, we proceed as follows.

$$\begin{align} \frac2\pi\int_0^{\pi/2}e^{4x\cos^2(\theta)}\,d\theta&=\frac2\pi\int_0^{\pi/2}\sum_{n=0}^\infty \frac{(4x)^n}{n!}\cos^{2n}(\theta)\,d\theta\\\\ &=\frac2\pi\sum_{n=0}^\infty \frac{(4x)^n}{n!}\int_0^{\pi/2}\cos^{2n}(\theta)\,d\theta\\\\ &=\frac2\pi\sum_{n=0}^\infty \frac{(4x)^n}{n!}\frac12 B\left(\frac12,n+\frac12\right)\\\\ &=\frac1\pi\sum_{n=0}^\infty \frac{(4x)^n}{n!}\frac{\Gamma(1/2)\Gamma(n+1/2}{\Gamma(n+1)}\\\\ &=\frac1\pi\sum_{n=0}^\infty \frac{\sqrt\pi(4x)^n}{(n!)^2}\Gamma(n+1/2)\\\\ &=\frac1\pi\sum_{n=0}^\infty \frac{\sqrt\pi(4x)^n}{(n!)^2}\frac{\sqrt \pi(2n-1)!!}{2^n}\\\\ &=\sum_{n=0}^\infty \frac{x^n\,(2n)!}{(n!)^3} \end{align}$$

as was to be shown!

We can use the previous result to show that $(3)$ in the OP is indeed correct. Note that we have

$$\begin{align} \overbrace{\int_0^{\pi/2}\cdots \int_0^{\pi/2}}^{m\,\text{ terms}} e^{4^mx\prod_{k=1}^m \cos^2{\theta_k}}\,d\theta_1\cdots\,d\theta_m&=\sum_{k=0}^\infty \frac{(4^{mk})x^k}{k!}\left(\int_0^{\pi/2}\cos^{2k}(\theta)\,d\theta\right)^m\\\\ &=\sum_{k=0}^\infty \frac{(4^{mk})x^k}{k!}\left(\frac{\pi}{2}\frac{(2k)!}{4^k(k!)^2}\right)^m\\\\ &=\left(\frac{\pi}{2}\right)^m\sum_{k=0}^\infty \frac{\left((2k)!\right)^m}{(k!)^{2m+1}}\,x^k \end{align}$$

After multiplying by $(2/\pi)^m$, we arrive at the result $(3)$!