# What Is The Limit Of The Sequence: $\frac{n^3}{{((3n)!)^\frac{1}{n}}}$

What (if exists) the $\lim \limits_{n\to \infty}\dfrac{n^3}{{((3n)!)^\frac{1}{n}}}$?

I have no idea where to begin. Maybe I could use the ratio test? Please try to keep it as elementary as possible because we are only in the beginning of the course.

Thanks a lot.

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A quick approach would be to use Stirling's approximation. – David Mitra Apr 12 '12 at 22:17
@DavidMitra as always, thanks for your great help. unfortunately we haven't gotten to logarithm during the course yet. – Anonymous Apr 12 '12 at 22:29
@Anonymous: the idea is to use $x!\sim \sqrt{2\pi x}\left(\dfrac xe\right)^x$ with $x=3n$. – Raymond Manzoni Apr 12 '12 at 22:31

We use the fact that if $(a_n)$ is a sequence of positive numbers and if $\lim\limits_{n\rightarrow\infty}{a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty}{\root n\of {a_n}}$ and the two limits are equal.

We apply this to the sequence with terms $a_n={n^{3n}\over (3n)!}$.

We have: $${a_{n+1}\over a_n} = {(n+1)^{3n+3}\over (3n+3)!}\cdot {(3n)!\over n^{3n}} ={(n+1)^3\over (3n+3)(3n+2)(3n+1) }\cdot {\Bigl({n+1\over n}\Bigr)^{3n} } \quad\buildrel{n\rightarrow\infty}\over\longrightarrow\quad{e^3\over27}.$$

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could you please show a little more in-depth why ${(n+1)^{3n+3}\over (3n+3)!}\cdot {(3n)!\over n^{3n}} ={(n+1)^3\over (3n+3)(3n+2)(3n+1) }\cdot {\Bigl({n+1\over n}\Bigr)^{3n} }$ – Anonymous Apr 12 '12 at 23:25
+1, this was the answer I was thinking about! – Aryabhata Apr 12 '12 at 23:26
@Anonymous $\bigl({n+1\over n}\bigr)^{3n} =\Bigl(\bigl(1+{1\over n}\bigr)^n\Bigr)^3$ tends to $e^3$. For the rational expression, expand the numerator and denominator and throw out negligible terms; you'll be left with $n^3\over27n^3$ (or divide every term in the numerator and denominator by $n^3$ after expanding everything). – David Mitra Apr 12 '12 at 23:29
@Anonymous This isn't what I suggested above, but gets the job done: $${(n+1)^3\over (3n+3)(3n+2)(3n+1)} ={n^3(1+{1\over n})^3\over 3n\bigl(1+{1\over n}\bigr)\cdot3n\bigl(1+{2\over 3n}\bigr)\cdot 3n\bigl(1+{1\over3 n}\bigr) }={ (1+{1\over n})^3\over 27\bigl(1+{1\over n}\bigr) \bigl(1+{2\over 3n}\bigr) \bigl(1+{1\over3 n}\bigr) }.$$ – David Mitra Apr 13 '12 at 0:20
$$\lim\limits_{n\rightarrow\infty}{n^3+2n\over 8n^3+5} =\lim\limits_{n\rightarrow\infty}{ {1\over n^3}(n^3+2n)\over {1\over n^3}(8n^3+5)} =\lim\limits_{n\rightarrow\infty}{{n^3\over n^3}+{2n\over n^3}\over {8n^3\over n^3}+{5\over n^3}}=\lim\limits_{n\rightarrow\infty}{ 1+ {2 \over n^2}\over8+{5\over n^3}} =\lim\limits_{n\rightarrow\infty}{ 1+0 \over 8+0 }={1\over8}.$$(It might be better to just factor out $n^3$ from the numerator and denominator. Then you get to fourth expression immediately.) – David Mitra Apr 13 '12 at 0:53

Ok, since David posted the approach I was thinking about, I wanted to see (just for fun) if there are other ways to find the limit in a completely basic fashion.

There is! but this is a bit long, so I might skip some details.

First we state a few Lemmas:

Lemma A:

If $x_n$ is a monotonically increasing sequence, and converges to $L$, then for all $n$, $L \ge x_n$.

Similarly if $y_n$ is a monotonically decreasing sequence, and $y_n$ converges to $R$, then for all $n$, $R \le y_n$.

Lemma B:

$x_n = \left(1 + \frac{1}{n}\right)^n$ is a monotonically increasing sequence.

Proof: Consider $n$ copies of $1 + \frac{1}{n}$ and $1$ copy of $1$ and apply $\text{AM} \ge \text{GM}$.

$\square$

Lemma C:

$y_n = \left(1 + \frac{1}{n}\right)^{n+1}$ is monotonically decreasing.

Proof: Consider $n$ copies of $1 - \frac{1}{n}$ and $1$ copy of $1$ and apply $\text{AM} \ge \text{GM}$.

$\square$

Note that Lemmas B and C can also be proven using Bernoulli's inequality, though that might not be considered basic (but surely can be considered elementary). These proofs (including the above AM,GM proofs) have appeared on this site before, if I remember correctly.

Combining the above Lemmas, leads us to:

Proposition D:

$$\left(1 + \frac{1}{n}\right)^{n+1} \ge e \ge \left(1 + \frac{1}{n}\right)^n$$

for all $n$.

Proof: $x_n$ and $y_n$ of Lemmas B and C, both converge to $e$ and we now have the inequality by Lemma A.

Note that, you can define $e$ this way too. First, $y_n$ is convergent as it is monotonically decreasing and bounded below. That means that $x_n$ is convergent too and to the same limit. Call that limit $e$.

Using binomial theorem, one can show that $x_n \lt 3$, and so $e \le 3$ (this can probably be found in most textbooks which define $e$ as the limit of $x_n$).

$\square$

Now we come to the main result:

Proposition E

For all $n \ge 4$, we have that

$$\left(\frac{n}{e}\right)^{n+8} \ge n! \ge \left(\frac{n}{e}\right)^{n-8}$$

Proof: By induction on $n$.

The base case is easily verifed using $e \le 3$.

Let $A_n = \left(\frac{n}{e}\right)^{n+8}$. Then

$$\frac{A_{n+1}}{A_n} = (n+1)\frac{\left(1 + \frac{1}{n}\right)^{n+8}}{e} \ge n+1$$

using Proposition D.

Thus if $A_n \ge n!$, then $A_{n+1} \ge (n+1)!$.

Similary, if $B_n = \left(\frac{n}{e}\right)^{n-8}$, the we can show that

$B_{n+1} \le (n+1)B_n$ and by induction, $B_{n+1} \le (n+1)!$.

$\square$

Now we can apply Proposition E to your sequence to squeeze your term and get that the limit is $\frac{e^3}{27}$.

I will leave that to the reader.

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the arithmetic mean of $1,2, \dots 3n$ should be $\frac{1+2+...+3n}{3n}$ and the geometric mean $\sqrt[3n]{1*2*3*...*3n}$? – Anonymous Apr 12 '12 at 22:43
@Anonymous: Actually I misread. I though that the denominator has the exponent of $\frac{1}{3n}$. This hint is not valid (at least the one I was thinking of). But there is another way without using logarithms (I think). Let me edit the answer. – Aryabhata Apr 12 '12 at 22:56
@Anonymous: I have edited with another answer, which is completely basic, but let me warn you, it is long. – Aryabhata Apr 13 '12 at 1:18
+1 Very nice!$\$ – David Mitra Apr 13 '12 at 1:50
@DavidMitra: Thanks! – Aryabhata Apr 13 '12 at 3:54

Stirling's approximation gives $$\lim_{n\to\infty}\frac1n(n!)^{1/n}=\frac1e\tag{1}$$ Using $(1)$ yields \begin{align} \lim_{n\to\infty}\frac{n^3}{{((3n)!)^{\frac1n}}} &=\lim_{n\to\infty}\frac{1}{27}\left(\frac{3n}{((3n)!)^{\frac{1}{3n}}}\right)^3\\ &=\frac{1}{27}\left(\lim_{n\to\infty}\frac{1}{3n}((3n)!)^{\frac{1}{3n}}\right)^{-3}\\ &=\frac{1}{27}\left(\frac1e\right)^{-3}\\ &=\frac{e^3}{27}\tag{2} \end{align}

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