GRE (5)

We approximate the factorials in the summation with Stirling's Approximation where $n! \approx \sqrt{2\pi n}\;(\frac{n}{e})^{n}$ . Therefore, it can be rewritten as $\begin{aligned} S &= \sum_{n=1}^\infty \frac{n!(2n)!}{(3n)!}x^{n} \\ & = \sum_{n=1}^\infty \frac{\sqrt{2\pi n}\;(\frac{n}{e})^{n} \sqrt{4\pi n}\;(\frac{2n}{e})^{2n}}{\sqrt{6\pi n}\;(\frac{3n}{e})^{3n}}x^{n} \\ & = \sum_{n=1}^\infty \frac{\sqrt{2}\; * 2\sqrt{\pi n}\;* 2^{2n} }{\sqrt{6}\;* 3^{3n}}x^{n} \\ & = \frac{2\sqrt{2\pi}}{\sqrt{6}} \sum_{n=1}^\infty \frac{4^{n}\sqrt{n}}{27^{n}}x^{n} \\ \end{aligned}$ This summation diverges if the exponential portion is greater than $1$ (the $\sqrt{n}$ is negligible compared to the exponentials). Therefore, $\boxed{7}$ is the answer because $4*7 = 28 > 27$ which leads the power series to diverge.

This is the computation for Aaghaz's solution. We use the ratio test for power series which says that a given series $\displaystyle \sum_0^{\infty} a_n$ is convergent if the limit $\displaystyle \lim_{n \to \infty}\Big| \frac{a_{n+1}}{a_n} \Big|$ is less than 1. The rest is computation: $\begin{aligned} &\lim_{n\to\infty} \Big| \frac{(n+1)! (2n+2)!}{(3n+3)!}x^{n+1} \cdot \frac{(3n!)}{(2n!)(n!)x^n} \Big| < 1\\ &|x| \lim_{n\to\infty} \Big|\frac{(2n+2)(2n+1)(n+1)}{(3n+3)(3n+2)(3n+1)} \Big| < 1. \\ \end{aligned}$ We do not need to expand the polynomials since they are similar degree and the limit goes to infinity, instead we only care about the ratio of the coefficients of the leading terms of the polynomials. Where $2\times 2\times 1 = 4$ and $3\times 3\times 3 =27$ , so the limit is $\frac{4}{27}$ and remembering this limit must be less than one for convergences gives: $\begin{aligned} &|x| \frac{4}{27} < 1\\ &|x| < \frac{27}{4} \\ &x<\frac{27}{4} = 6.75. \end{aligned}$ So the nearest integer that causes divergence is $\boxed{7}$ .

For the power series to diverge, the ratio of the (n+1)st term to the (n)th term should be greater than 1 as n approaches infinity.......solving this, we get x > 6.75......hence minimum positive integral x is 7......