# Convergence of the series $\sum\limits_{n=1}^{\infty} \frac{(n!)^3 x^n}{n(3n)!}$

I am studying the convergence of the following series: $$\sum\limits_{n=1}^{\infty} \frac{(n!)^3 x^n}{n(3n)!}$$.

I have proceeded as follows:

\begin{align*} \left|\frac{\frac{(n+1)!^3x^{n+1}}{(n+1)(3(n+1))!}}{\frac{(n!)^3 x^n}{n(3n)!}}\right|&=\left|\frac{(n+1)!^3x^{n+1}}{(n+1)(3(n+1))!}\cdot\frac{n(3n)!}{(n!)^3 x^n}\right|=\left|\frac{(n+1)^3\cdot x\cdot n}{(n+1)(3n+3)(3n+2)(3n+1)}\right|=\left|\frac{n^4\left(1+\frac{1}{n}\right)^3\cdot x}{27n^4\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(1+\frac{2}{3n}\right)\left(1+\frac{1}{3n}\right)}\right|\\ &=\left|\frac{\left(1+\frac{1}{n}\right)^3\cdot x}{27\left(1+\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(1+\frac{2}{3n}\right)\left(1+\frac{1}{3n}\right)}\right|\xrightarrow[]{n\to\infty}\left|\frac{x}{27}\right|=\frac{|x|}{27} \end{align*} so if $$\frac{|x|}{27}<1\Leftrightarrow |x|<27\Leftrightarrow -27 the series is absolutely convergent by the ratio test and if $$\frac{|x|}{27}>1\Leftrightarrow |x|>27\Leftrightarrow x>27\text{ or }x<-27$$ then the series diverges since it cannot be that $$\lim\limits_{n\to\infty}a_n=0$$ because in that case it should be $$\lim\limits_{n\to\infty}|a_n|=0$$, a contradiction.

If $$x=27$$ then $$a_n=\frac{(n!)^3 (27)^n}{n(3n)!}$$ so \begin{align*} n\left(\frac{a_n}{a_{n+1}}-1\right)&=n\left(\frac{(n!)^3 (27)^n}{n(3n)!}\cdot \frac{(n+1)(3n+3)!}{(n+1)!^3 (27)^{n+1}}-1\right)\\ &=n\left(\frac{(3n+1)(3n+2)(3n+3)}{27n(n+1)^2}-1\right)=n\left(\frac{27n^3(1+\frac{1}{3n})(1+\frac{2}{3n})(1+\frac{1}{n})}{27n^3(1+\frac{1}{n})^2}-1\right)\\ &=n\left(\frac{(1+\frac{1}{3n})(1+\frac{2}{3n})}{1+\frac{1}{n}}-1\right)=n\left(\frac{1+\frac{1}{n}+\frac{2}{9n^2}}{1+\frac{1}{n}}-1\right)\\ &=n\left(\frac{\frac{2}{9n^2}}{1+\frac{1}{n}}\right)\leq\frac{2}{9}<1\ \forall n\in\mathbb{N}^+ \end{align*} so the series is divergent by Raabe's test.

If $$x=-27$$ then $$a_n=\frac{(n!)^3 (-27)^n}{n(3n)!}=(-1)^n\frac{(n!)^3 (27)^n}{n(3n)!}$$ and I haven't been able to prove that it either converges or diverges in this case. Numerical inspection suggests that the series diverges, and I think the best way to prove this is to show that $$\lim\limits_{n\to\infty}\frac{(n!)^3 (27)^n}{n(3n)!}\neq 0$$ and I have tried to do so firstly by using the ratio test, which is inconclusive in this case, and then by rewriting $$\frac{(n!)^3 (27)^n}{n(3n)!}\geq\frac{(n!)^3 (27)^n}{n[(1\cdot 2\cdot\ldots\cdot n)((n+1)\cdot (n+2)\cdot\ldots\cdot 2n)(2n+1)\cdot\ldots\cdot 3n]}$$ and trying to prove that this expression is $$\geq K$$, where $$K>1$$, but I haven't been able to do so.

Thus, I would be grateful if someone would tell me how to prove that this limit does not converge to $$0$$. Thanks

NOTE: in the real analysis book I am reading Stirling's approximation for $$n!$$ is stated and proved three chapters after this problem, so I think one ought to be able to solve this without using it and I would like to know precisely how to do this.

• How about using Stirling's approximation for $n!$? Nov 16, 2023 at 11:59
• @VIVID in the real analysis book I am reading Stirling's approximation for $n!$ is stated and proved three chapters after this problem, so I think one ought to be able to solve this without using it. Thanks for the hint, though. Nov 16, 2023 at 12:10
• you can do it easily by using the easy-to-prove inequalities $n! >(n/e)^n$ and $n!< Cn (n/e)^n$ with the first following from the Taylor series of $e^n$ which is greater than any term so in particular than the $n$ th term which is $n^n/n!$; for the other inequality show that the terms have a maximum at $n^n/n!$ and the terms from $k \ge 10n$ say are exponentially small so $e^n \le 10n \times (n^n/n!)(1 +o(1)) \le 20n \times n^n/n!$ Nov 16, 2023 at 15:44
• @Conrad Perhaps I'm doing bad algebra, but does that not give a lower bound of $1/(3 C n^2) \to 0$ as $n \to \infty$? Nov 16, 2023 at 15:51
• Indeed the numerator and denominator really are about the same size, so one can't get away with looser bounds than the real kind of size $\sqrt{n} (n/e)^n$ if going the route of bounding $n!$ individually. Nov 16, 2023 at 16:16

Stirling's approximation is indeed a fine way to show that $$\lim_{n \to \infty} \frac{(n!)^3 27^n}{n (3n)!} = \frac{2 \pi}{\sqrt{3}}$$ but here's a way to get this lower bound without it. (Whether or not anything of the sort is the intended approach in the textbook I cannot know---I don't know what has been discussed at this point in your book.)
Let us start by rewriting $$n! = \Gamma(n + 1)$$ and $$(3n)! = \Gamma(3n + 1) = 3n \Gamma(3n)$$, with which we get $$\frac{(n!)^3 27^n}{n (3n)!} = \frac{\Gamma(n + 1)^3 27^n}{3n^2 \Gamma(3n)}.$$
The triplication formula for Gamma says that $$\Gamma(3n) = (2 \pi)^{-1} 3^{3n - 1/2} \Gamma(n) \Gamma(n + 1/3) \Gamma(n + 2/3),$$ so our expression becomes $$\frac{ 2 \pi \Gamma(n + 1)^3 27^n}{3n^2 3^{3n - 1/2} \Gamma(n) \Gamma(n + 1/3) \Gamma(n + 2/3)}.$$ Cleaning this up we have $$\frac{2 \pi \Gamma(n + 1)^3}{\sqrt{3} n^2 \Gamma(n) \Gamma(n + 1/3) \Gamma(n + 2/3)}.$$ One of the Gammas up top is handled by $$\Gamma(n + 1) = n \Gamma(n)$$, so we are left with $$\frac{2 \pi \Gamma(n + 1)^2}{\sqrt{3} n \Gamma(n + 1/3) \Gamma(n + 2/3)}.$$ This is where Gautschi's inequality $$x^{1 - s} < \frac{\Gamma(x + 1)}{\Gamma(x + s)}$$ for $$x > 0$$ and $$0 < s < 1$$ real comes in: if we apply this with $$s = 1/3$$ and $$s = 2/3$$ we get $$\frac{2 \pi \Gamma(n + 1)^2}{\sqrt{3} n \Gamma(n + 1/3) \Gamma(n + 2/3)} > \frac{2 \pi n^{1 - 1/3} n^{1 - 2/3}}{\sqrt{3} n} = \frac{2 \pi}{\sqrt{3}}.$$ Hence $$\frac{(n!)^3 27^n}{n (3n)!} > \frac{2 \pi}{\sqrt{3}}.$$
Note by the way that this is a pretty delicate limit: if we use worse bounds in the place were we apply Gautschi's inequality we just miss the mark. What I mean by this is: since $$\Gamma(x)$$ is increasing for $$x > 2$$ (say), we could've tried something naive like $$\Gamma(n + 1/3) < \Gamma(n + 1)$$ and cancel the Gamma terms this way. This would leave the factor of $$n$$ in the denominator, however, and give us a lower bound of $$0$$ in the limit, which isn't good enough for our purposes.