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 $\frac{|x|}{27}<1\Leftrightarrow |x|<27\Leftrightarrow -27<x<27$ 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.

  • $\begingroup$ How about using Stirling's approximation for $n!$? $\endgroup$
    – VIVID
    Nov 16, 2023 at 11:59
  • $\begingroup$ @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. $\endgroup$
    – lorenzo
    Nov 16, 2023 at 12:10
  • $\begingroup$ 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 $e^n \le 10n \times (n^n/n!)(1 +o(1)) \le 20n \times n^n/n!$ $\endgroup$
    – Conrad
    Nov 16, 2023 at 15:44
  • $\begingroup$ @Conrad Perhaps I'm doing bad algebra, but does that not give a lower bound of $1/(3 C n^2) \to 0$ as $n \to \infty$? $\endgroup$ Nov 16, 2023 at 15:51
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 16, 2023 at 16:16

1 Answer 1


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 tricks I will use are Gautschi's inequality, which is a consequence of the log-convexity of the Gamma function, and the triplication formula, which doesn't require Stirling (though using Stirling does give a nice argument for the general multiplication theorem).

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.


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