# $L=\lim_{n\to \infty}\frac{1}{\sqrt[n]{n!}},$ then which of the value of L is possible. [closed]

[CSIR-UGC NET Examination, 2017 June session]

$L=\lim_{n\to \infty}\frac{1}{\sqrt[n]{n!}}$. Then

(1)$L=0$

(2)$L=1$

(3)$0<L<1$

(4) $L=\infty$

Let $x_n=\frac{1}{\sqrt[n]{n!}}$.

Taking logarithm on both sides,

$\log(x_n)=\frac{1}{n}\sum_{k=1} log(\frac{1}{k})$

Using the Cauchy's first theorem on limits,$\lim_{n\to \infty}\log(\frac{1}{n})=-\infty$. So $\lim_{n\to \infty}x_n=0$. Am I correct? Please suggest some short methods. Please note that typing error had occured in the main title. I corrected it. I am apologising for the error.

## closed as off-topic by Yves Daoust, Hurkyl, José Carlos Santos, Jack D'AurizioNov 19 '17 at 21:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hurkyl, José Carlos Santos, Jack D'Aurizio
If this question can be reworded to fit the rules in the help center, please edit the question.

• Isn't it obvious that this is a case of $1/\infty$ ? – Yves Daoust Nov 19 '17 at 11:19
• @YvesDaoust This result confuses me to apply this, math.stackexchange.com/questions/28348/… – Unknown x Nov 19 '17 at 11:20
• I didn't think about method given by Find_X.:( – Unknown x Nov 19 '17 at 11:22
• @YvesDaoust His answer is wrong. – Unknown x Nov 19 '17 at 11:34
• Silently chaging the question is very bad practice. It makes all contributions nonsensical. – Yves Daoust Nov 19 '17 at 11:43

Note that $0\le \dfrac{1}{\sqrt {n!}}\le \dfrac{1}{\sqrt n}$

Since $\dfrac{1}{\sqrt n}\to 0$ as $n\to \infty$ by Squeeze Theorem we have $L=0$

• @ManeeshNarayanan; Why did you change the question by editing it?You could have asked a new one.This is not done – Learnmore Nov 19 '17 at 11:37
• Title has typing error. – Unknown x Nov 19 '17 at 11:38
• Why didn't you read the question completely?. – Unknown x Nov 19 '17 at 11:40
• @ManeeshNarayanan: the editing history clearly shows that you fixed the expression. You should just apologize. – Yves Daoust Nov 19 '17 at 11:46
• @YvesDaoust. I haven't changed the body of the question. There was error in the title. I was not aggressive.@Find_x If you hurt my comments, I am apologising for that. sorry. – Unknown x Nov 19 '17 at 11:57

Use that if $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}$ exist and $a_n\geq 0$ then $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=\lim_{n\to\infty} \sqrt[n]{a_n}$. So: \begin{align} \lim_{n\to \infty}\sqrt[n]{\frac{1 } {n! } }=\lim_{n\to \infty}\frac{n!} {(n+1)!}=0 \end{align}

Note that for $x>0$ and $n\in \Bbb N$ we have $e^x = \displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!} \geq \frac{x^n}{n!}$.

Setting $x=n$, we get $e^n \geq \frac{n^n}{n!}$, hence $e \geq \frac{n}{\sqrt[n]{n!}}$, so $\frac{e}{n} \geq \frac{1}{\sqrt[n]{n!}} \geq 0$, thus $\lim_{n\to \infty}\frac{1}{\sqrt[n]{n!}} = 0$ by the squeeze theorem.

Let's try to find $$\lim\limits_{n\rightarrow\infty}(n!)^{\frac{1}{n}}$$

By the Stirling approximation this is equal to

$$\lim\limits_{n\rightarrow\infty}\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)^{\frac{1}{n}}=\lim\limits_{n\rightarrow\infty}\left(\sqrt{2\pi n}\right)^{\frac{1}{n}}\left(\frac{n}{e}\right)$$

which obviously diverges. Thus the correct answer is indeed $L=0$