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I was trying to solve the limit:

$$\lim_{n \to \infty} \sqrt[n]{\frac{(3n)!}{(5n)^{3n}}}$$

By using the root's criterion for limits (which is valid in this case, since $b_n$ increases monotonically):

$$L= \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$

Now I realise using Sterling's formula would make everything easier, but my first approach was simplifying the factorial after applying the criterion I mentioned before. So, after a few failed attempts I looked it up on Mathematica and it said that $\frac{(3(n+1))!}{(3n)!}$ (which is one of the fractions you have to simplify) equals $3(n+1)(3n+1)(3n+2)$. Since I can't get there myself I want to know how you would do it.

Just so you can correct me, my reasoning was:

$$\frac{(3(n+1))!}{(3n)!} = \frac{3\cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdot (...) \cdot 3 \cdot (n+1)}{3 \cdot 1 \cdot 3 \cdot 2 \cdot 3 \cdot 3 \cdot (...) \cdot 3 \cdot n } = $$ $$= \frac{3^n(n+1)!}{3^{n}n!} = \frac{(n+1)!}{n!} = n+1$$

Which apparently isn't correct. I must have failed at something very silly. Thanks in advance!

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    $\begingroup$ $(3(n+1)!)$ is the product of all the natural numbers from $1$ to $3(n+1)$, and $3\cdot1\cdot 3\cdot 2\ldots\cdot 3(n+1)$ is the product of the numbers that are multiples of $3$. $\endgroup$ – ajotatxe Jan 22 '17 at 19:08
  • $\begingroup$ @Manuel Also you can apply Stirling's approximation. $\endgroup$ – Behrouz Maleki Jan 22 '17 at 19:11
  • $\begingroup$ @ajotatxe I don't have (3(n+1)!), but (3(n+1))! (the factorial includes everything between the parentheses). $\endgroup$ – mar Jan 22 '17 at 19:17
  • $\begingroup$ This question is somewhat similar: Need to simplify ratio test: $\frac{(2(k+1))!}{(2k)!} $ - it might also be useful for you. I found it using Approach0. $\endgroup$ – Martin Sleziak Jan 23 '17 at 6:49
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By Stirling's inequality the answer is clearly $\left(\frac{3}{5e}\right)^3$. To prove it, you may notice that by setting $$ a_n = \frac{(3n)!}{(5n)^{3n}} $$ you have: $$ \frac{a_{n+1}}{a_n} = \frac{(3n+3)(3n+2)(3n+1)(5n)^{3n}}{(5n+5)^{3n+3}} = \frac{\frac{3n+3}{5n+5}\cdot\frac{3n+2}{5n+5}\cdot\frac{3n+1}{5n+5}}{\left(1+\frac{1}{n}\right)^{3n}}\to\frac{\left(\frac{3}{5}\right)^3}{e^3}$$ as $n\to +\infty$.

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    $\begingroup$ I don't want to know the answer to the limit, but how I can go from a reason of factorials to the polynomial Mathematica gave me. But thanks, I'll use your answer to check if I got it right, haha. $\endgroup$ – mar Jan 22 '17 at 19:15
  • $\begingroup$ @Manuel: the reason is simple: if you have a positive sequence such that $\frac{a_{n+1}}{a_n}\to c$, then $\sqrt[n]{a_n}\to c$, too. $\endgroup$ – Jack D'Aurizio Jan 22 '17 at 19:16
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    $\begingroup$ And $$\frac{(3(n+1))!}{(3n)!}=\frac{(3n+3)!}{(3n)!}=(3n+3)(3n+2)(3n+1).$$ $\endgroup$ – Jack D'Aurizio Jan 22 '17 at 19:18
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    $\begingroup$ Oh, that makes sense. Why did I not see it before? Thanks a lot, Jack. $\endgroup$ – mar Jan 22 '17 at 19:19
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I thought it might be useful to present an approach that relies on elementary tools only. To that end, we proceed.

First, we write

$$\begin{align} \frac{1}{n}\log((3n!))&=\frac1n\sum_{k=1}^{3n}\log(k)\\\\ &=\left(\frac1n\sum_{k=1}^{3n}\log(k/n)\right)+3\log(n) \tag 1 \end{align}$$

Now, note that the parenthetical term on the right-hand side of $(1)$ is the Riemann sum for $\int_0^3 \log(x)\,dx=3\log(3)-3$.

Using $(1)$, we have

$$\begin{align} \lim_{n\to \infty}\sqrt[n]{\frac{(3n)!}{(5n)^{3n}}}&=\lim_{n\to \infty}e^{\frac1n \log((3n)!)-3\log(5n)}\\\\ &=\lim_{n\to \infty}e^{\left(\frac1n\sum_{k=1}^{3n}\log(k/n)\right)+3\log(n)-3\log(5n)}\\\\ &=\lim_{n\to \infty}e^{\left(\frac1n\sum_{k=1}^{3n}\log(k/n)\right)-3\log(5)}\\\\ &= e^{3\log(3)-3-3\log(5)}\\\\ &=\left(\frac{3}{5e}\right)^3 \end{align}$$

as expected!

Tools used: Riemann sums and the continuity of the exponential function.

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$$ (3(n+1))! \neq 3 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 4 \cdots 3 \cdot (n+1) $$ To make it clear what the problem is, let's write the right-hand side with brackets: $$ (3 \cdot 2) \cdot (3 \cdot 3) \cdot (3 \cdot 4) \cdots (3 \cdot (n+1)) $$ That's just multiplying all the positive multiples of 3 less than $ 3(n+1) $ together; the factorial is defined as multiplying together all positive integers less than $ 3(n+1) $. So the correct expansions are \begin{align*} (3(n+1))! &= 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots (3n-2) \cdot (3n-1) \cdot 3n \cdot (3n+1) \cdot (3n+2) \cdot 3(n+1) \\ (3n)! &= 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots (3n-2) \cdot (3n-1) \cdot 3n \end{align*} which clearly have the quotient Mathematica gave you.

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  • $\begingroup$ Thanks a lot, I noticed what you just said reading Jack's response. I haven't used the factorial function a lot and now I see I didn't completely understand it (or maybe I'm especially slow today). $\endgroup$ – mar Jan 22 '17 at 19:35
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My favorite elementary inequality for $n!$: $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $ $(n/e)^n < n! < (n/e)^{n+1} $.

Taking n-th roots, $n/e < (n!)^{1/n} < (n/e)^{1+1/n} $.

Therefore $\sqrt[n]{\lfrac{(3n)!}{(5n)^{3n}}} =\lfrac{((3n)!)^{1/n}}{(5n)^{3}} >\lfrac{((3n/e)^{3n})^{1/n}}{(5n)^{3}} =\lfrac{(3n/e)^{3}}{(5n)^{3}} =\lfrac{27n^3}{125e^3n^{3}} =\lfrac{27}{125e^3} $.

Arguing the other way, $\sqrt[n]{\lfrac{(3n)!}{(5n)^{3n}}} =\lfrac{((3n)!)^{1/n}}{(5n)^{3}} <\lfrac{((3n/e)^{3n+1})^{1/n}}{(5n)^{3}} =\lfrac{(3n/e)^{3}(3n/e)^{1/n}}{(5n)^{3}} =\lfrac{27n^3(3n/e)^{1/n}}{125e^3n^{3}} =\lfrac{27}{125e^3}(3n/e)^{1/n} $. Since, as $n \to \infty$, $n^{1/n} \to 1 $ and $a^{1/n} \to 1 $ for any $a > 0$, $(3n/e)^{1/n} \to 1 $ so the limit is $\lfrac{27}{125e^3} $.

A more general result:

$\sqrt[n]{\lfrac{(an)!}{(bn)^{an}}} =\lfrac{((an)!)^{1/n}}{(bn)^a} \gt \lfrac{((an/e)^{an})^{1/n}}{(bn)^a} =\lfrac{(an/e)^a}{(bn)^a} =\lfrac{a^a n^a}{b^ae^a} =\lfrac{a^a}{b^ae^a} =\left(\lfrac{a}{be}\right)^a $.

The other way goes exactly as above, so that $\lim_{n \to \infty} \sqrt[n]{\lfrac{(an)!}{(bn)^{an}}} =\left(\lfrac{a}{be}\right)^a $.

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  • $\begingroup$ I get the part where you say this: $\frac{((an)!)^{1/n}}{(bn)^a} \gt \frac{((an/e)^{an})^{1/n}}{(bn)^a}$ but why would that imply that $(a/be)^a$ is the result of the limit? $\endgroup$ – mar Jan 22 '17 at 20:25
  • $\begingroup$ The $n^a$ terms cancel out. $\endgroup$ – marty cohen Jan 22 '17 at 23:13
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    $\begingroup$ I made your fractions a bit bigger so that they are easier to read. Hope you don't mind, but feel free to revert if you don't like it. =) $\endgroup$ – user21820 Jan 23 '17 at 3:49
  • $\begingroup$ I don't mind editing as long as the meaning isn't changed. $\endgroup$ – marty cohen Jan 23 '17 at 4:12

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