[CSIR-UGC NET Examination, 2017 June session]

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




(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'Aurizio Nov 19 '17 at 21:49

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  • 1
    $\begingroup$ Isn't it obvious that this is a case of $1/\infty$ ? $\endgroup$ – Yves Daoust Nov 19 '17 at 11:19
  • $\begingroup$ @YvesDaoust This result confuses me to apply this, math.stackexchange.com/questions/28348/… $\endgroup$ – Unknown x Nov 19 '17 at 11:20
  • $\begingroup$ I didn't think about method given by Find_X.:( $\endgroup$ – Unknown x Nov 19 '17 at 11:22
  • $\begingroup$ @YvesDaoust His answer is wrong. $\endgroup$ – Unknown x Nov 19 '17 at 11:34
  • 1
    $\begingroup$ Silently chaging the question is very bad practice. It makes all contributions nonsensical. $\endgroup$ – 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 $L=0$

  • $\begingroup$ @ManeeshNarayanan; Why did you change the question by editing it?You could have asked a new one.This is not done $\endgroup$ – Learnmore Nov 19 '17 at 11:37
  • $\begingroup$ Title has typing error. $\endgroup$ – Unknown x Nov 19 '17 at 11:38
  • $\begingroup$ Why didn't you read the question completely?. $\endgroup$ – Unknown x Nov 19 '17 at 11:40
  • $\begingroup$ @ManeeshNarayanan: the editing history clearly shows that you fixed the expression. You should just apologize. $\endgroup$ – Yves Daoust Nov 19 '17 at 11:46
  • $\begingroup$ @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. $\endgroup$ – 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 $L=0$


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