# How to solve this limit problem?-$\lim_{n\to \infty}\ \left(\frac{\ n!}{(mn)^n}\right)^{\frac{1}{n}}$

I need to find the value of-

$$\lim_{n\to \infty}\ \left(\frac{\ n!}{(mn)^n}\right)^{\frac{1}{n}}$$

where $m {\in} R$

I don't know how to even start. Would someone explain it step by step, also which type of indeterminate form is this? Is there a simpler to solve this ? i.e. without any high mathematics theorem etc.?

• HInt: If $a_n$ is a positive sequence and $\lim a_n^{1/n}$ exists, then $\lim a_n^{1/n} = \lim \frac{a_{n+1}}{a_n}$ :) – r9m Apr 1 '16 at 14:21
• It should be $m\in\mathbb{R}$, $m>0$. For $m<0$ the limit doesn't exist. – egreg Apr 1 '16 at 14:52
• First write it in the form $(\frac{n!}{(mn)^n})^{1/n} = \frac{1}{m} (\frac{n!}{n^n})^{1/n}$ so that the factor $m$ is a less of a distraction, then look at $\frac{(n!)^{1/n}}{n}$. – jim Apr 1 '16 at 18:05

Use Stirling's approximation for $n!$

$n!\approx \sqrt{2\pi n}( \frac ne)^n$

$$\lim_{n\to \infty} \left(\frac{\sqrt{2\pi n}}{(em)^n}\right)^{\frac 1n}$$

$$\frac {1}{em} \lim_{n\to \infty} (2\pi n)^{\frac 1{2n}}$$

Let the above limit (excluding constant) be $L$

$$\log L= \lim_{n\to \infty}\frac{\log (2\pi n)}{2n}$$

L'Hopital's rule here

• well i haven't studied that approximation but still thanks. – user323082 Apr 1 '16 at 14:27
• @Bubububu_it seems that the approximation didn't work here. Can you tell me when it is applicable? – user323082 Apr 1 '16 at 14:42
• @XBOveRLorD What do you mean by doesn't work? – lEm Apr 1 '16 at 14:46
• the answer is 1/em – user323082 Apr 1 '16 at 14:47
• @XBOveRLorD when $n>1$, the approximation is really accurate. en.wikipedia.org/wiki/Stirling%27s_approximation – lEm Apr 1 '16 at 14:48

You don't need Stirling.

$\left(\dfrac{\ n!}{(mn)^n}\right)^{\frac{1}{n}} =\dfrac{(n!)^{1/n}}{mn}$ so all you need is that $\dfrac{(n!)^{1/n}}{n} \to \dfrac{1}{e}$ to get that the limit is $\dfrac1{em}$.

$\dfrac{(n!)^{1/n}}{n} \to \dfrac{1}{e}$ follows from $\left(\dfrac{n}{e}\right)^n < n! <\left(\dfrac{n}{e}\right)^{n+1}$. These, in turn, can be proved by induction from $\left(1+\dfrac1{n}\right)^n < e <\left(1+\dfrac1{n}\right)^{n+1}$.

• Marty, do you know/see another way proving your 2th last inequality than using induction or Stirling? Is there another handy tool one can try? – Imago Apr 1 '16 at 15:29
• Not easily. This kind of inequality, when you do not use logs or calculus or limits, can usually be proved only by induction. Those inequalities bounding n! are elementary and can be very useful (as shown here). – marty cohen Apr 1 '16 at 15:41
• I was afraid that's the case :/, I always try to get better at manipulating terms, cause I find after some point inductions, L'Hospital or Stirling doesn't add anything to one's knowledge/understanding and thus become more and more the boundary of problems one can solve. – Imago Apr 1 '16 at 15:47
• A fairly crude estimate for $n!$ is adequate because of the effect of the $n$th root opertion. – DanielWainfleet Apr 1 '16 at 18:04

let $a_n = \frac {n!}{(mn)^n}$. by the root-frac analogy we know that for every $a_n$ if the limit of the series $b_n = \frac{a_{n+1}}{a_n}$ exist it is equal to the limit of the series $c_n = \sqrt[n]{a_n}$. so lets calculate the limit of the series $b_n = \frac{a_{n+1}}{a_n}$

$b_n = \frac{a_{n+1}}{a_n} = \frac {(n+1)!}{(m((n+1)))^{(n+1)}} : \frac {n!}{(mn)^n} = m^{-1}(\frac{n}{n+1})^n$ we know that the limit of $(\frac{n}{n+1})^n$ (n goes to infinity) is $e^{-1}$ then the limit of $b_n$ (since m is a constant different than 0) is $m^{-1}e^{-1} = \frac{1}{me}$

final result: $\frac{1}{me}$

$$\lim_{n\to\infty}\left(\frac{n!}{(mn)^n}\right)^{\frac{1}{n}}=\lim_{n\to\infty}\exp\left[\ln\left(\left(\frac{n!}{(mn)^n}\right)^{\frac{1}{n}}\right)\right]=$$ $$\lim_{n\to\infty}\exp\left[\frac{\ln\left(\frac{n!}{(mn)^n}\right)}{n}\right]=\exp\left[\lim_{n\to\infty}\frac{\ln\left(\frac{n!}{(mn)^n}\right)}{n}\right]=$$

Apply l'Hôpital's rule:

$$\exp\left[\lim_{n\to\infty}\frac{\psi^{(0)}(1+n)-\ln(mn)-1}{1}\right]=$$ $$\exp\left[\lim_{n\to\infty}\left(\psi^{(0)}(1+n)-\ln(mn)-1\right)\right]=$$ $$\exp\left[\left(0-\ln(m)-1\right)\right]=\exp\left[-\ln(m)-1\right]=\frac{1}{me}$$

• To find this limit $\lim_{n\to\infty}\left(\psi^{(0)}(1+n)-\ln(mn)-1\right)$, notice:

$$\lim_{n\to\infty}\left(\psi^{(0)}(1+n)-\ln(mn)-1\right)=$$ $$\lim_{n\to\infty}\left(\psi^{(0)}(1+n)-\ln(mn)\right)-1=$$ $$\lim_{n\to\infty}\ln\left(\frac{\exp\left[\psi^{(0)}(1+n)\right]}{mn}\right)-1=\ln\left(\frac{1}{m}\right)-1$$

• please explain the l'hospital rule step you used – user323082 Apr 1 '16 at 14:38
• @XBOveRLorD Take the derivative of the denominator en nominator and you'll find what I found – Jan Apr 1 '16 at 14:39
• i dont know about that psi or whatever function – user323082 Apr 1 '16 at 14:40
• $\psi^{(n)}(x)$ is the derivative of the digamma function (mathworld.wolfram.com/PolygammaFunction.html) – Jan Apr 1 '16 at 14:43
• is there any other way to solve the limit? – user323082 Apr 1 '16 at 14:45