5

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.

Aryabhata
  • 82,206
Anonymous
  • 2,388

3 Answers3

15

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

David Mitra
  • 74,748
  • 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
    @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
  • Sorry, but how do I develop this expression ${(n+1)^{3n+3}\over (3n+3)!}\cdot {(3n)!\over n^{3n}}=?$ – Anonymous Apr 12 '12 at 23:38
  • @Anonymous$${ {(n+1)^{3n+3}\over(3n+3)!}\over{n^{3n}\over(3n)!} }={(n+1)^{3n+3}\over (3n+3)!}\cdot{(3n)!\over n^{3n}}={\color{darkgreen}{(n+1)^{3n}}(n+1)^3\over (3n+3)(3n+2)(3n+1)\color{maroon}{(3n)!}}\cdot{\color{maroon}{(3n)!}\over \color{darkgreen}{n^{3n}}}.$$ – David Mitra Apr 12 '12 at 23:41
  • @David Mitra You're amazing, thank you so much! :-) – Anonymous Apr 12 '12 at 23:51
  • @DavidMitra last thing actually, could you please show that $\frac{(n+1)^3}{(3n+3)(3n+2)(3n+1)}=\frac{n^3}{27n^3}$? – Anonymous Apr 12 '12 at 23:53
  • @Anonymous they aren't actually equal, but can you see that for the purposes of taking the limit they are essentially the same? – Ragib Zaman Apr 12 '12 at 23:59
  • @Ragib Zaman why for the purpose of taking the limit they are essentially the same? – Anonymous Apr 13 '12 at 00:16
  • 1
    @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 00:20
  • @David Mitra that was really nice :-) – Anonymous Apr 13 '12 at 00:25
  • @David Mitra I still would like to understand the "trick" you made in your previous argument; I expanded the expression $\frac{(n+1)^3}{(3n+3)(3n+2)(3n+1)}$ and got that it's equal to: $\frac{n^3+3n^2+3n+1}{27n^3+54n^2+33n+6}$ which isn't obviously $\frac{n^3}{27n^3}$; what are these "negligible terms" and why are they negligible? – Anonymous Apr 13 '12 at 00:28
  • @Anonymous It's the highest powers that count when taking a limit at infinity of a rational expression. Divide every term by $n^3$ (or, equivalently, factor out $n^3$ upstairs and downstairs). Once you've done this my remark about "negligible terms" should seem less cryptic... – David Mitra Apr 13 '12 at 00:35
  • @David Mitra sorry, but I want to be sure that I understand you as I'm sure it will happen a lot during the course; let's take a simpler expression: $\frac{n^3+2n}{8n^3+5}$; how do I divide every term by $n^3$ and how does it help me? – Anonymous Apr 13 '12 at 00:43
  • 1
    $$\lim\limits_{n\rightarrow\infty}{n^3+2n\over 8n^3+5} =\lim\limits_{n\rightarrow\infty}{ {1\over n^3}(n^3+2n)\over {1\over n^3}(8n^3+5)} =\lim\limits_{n\rightarrow\infty}{{n^3\over n^3}+{2n\over n^3}\over {8n^3\over n^3}+{5\over n^3}}=\lim\limits_{n\rightarrow\infty}{ 1+ {2 \over n^2}\over8+{5\over n^3}} =\lim\limits_{n\rightarrow\infty}{ 1+0 \over 8+0 }={1\over8}.$$(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 00:53
  • @David Mitra I can't stress how much I appreciate your time and willingness to help; you helped me a lot, THANKS!!! – Anonymous Apr 13 '12 at 01:00
  • @Anonymous You're very welcome; glad to help :) – David Mitra Apr 13 '12 at 01:01
11

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 $n$, $R \le y_n$.

Proof: Left to the reader.

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.

Aryabhata
  • 82,206
  • the arithmetic mean of $1,2, \dots 3n$ should be $\frac{1+2+...+3n}{3n}$ and the geometric mean $\sqrt[3n]{123...3n}$? – 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 01:18
  • @DavidMitra: Thanks! – Aryabhata Apr 13 '12 at 03:54
  • @Aryabhata First of all, thank you very much for the very in-depth proof, I really appreciate it! In Lemma B did you mean to write the following inequality: $\frac{1+n(n+\frac{1}{n})}{1+n} > \sqrt[n+1]{(1+\frac{1}{n})^n}$? – Anonymous Apr 13 '12 at 08:29
  • @Anonymous: Yes. And then manipulate that to show that $x_{n+1} \gt x_n$. – Aryabhata Apr 13 '12 at 14:40
6

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

Alternate Proof of $\boldsymbol{(1)}$ $$ \begin{align} \log\left(\lim_{n\to\infty}\frac1n(n!)^{1/n}\right) &=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\log\left(\frac kn\right)\\ &=\int_0^1\log(x)\,\mathrm{d}x\\[9pt] &=-1\tag3 \end{align} $$ which is equivalent to $(1)$.

robjohn
  • 345,667