The limit superior of R given as lim sup of $\frac{1}{\left| \left(\frac{1}{n!}\right)^\frac{1}{n} \right|}$ as n approaches infinity

Question

What is the limit superior of $\dfrac{1}{\left| \left(\frac{1}{n!}\right)^\frac{1}{n} \right|}$ as n approaches infinity?

So far I have been able to establish that $n≤n!$ for $n≥1$ which the gives that $\frac1{n!}≤\frac1n$ from there I used the sandwich theorem on $0≤\frac1{n!}≤\frac1n$ to reach the conclusion that $\frac1{n!}$ tends towards zero as $n$ approaches infinity. I am now stuck on how to deal with the exponent $\frac1n$.

So far I have been able to establish that $n \leq n!$ for $n \geq 1$ which the gives that $\frac{1}{n!} \leq \frac{1}{n}$ from there I used the sandwich theorem on $0 \leq \frac{1}{n!} \leq \frac{1}{n}$ to reach the conclusion that $\frac{1}{n!}$ tends towards zero as n aprochies infinity. I am now stuck on how to deal with the exponent $\frac{1}{n}$. — 1721, Sep 12 '19 at 16:41
See also: How to show $(1/n!)^{1/n}$ goes to $0$ as $n$ goes to infinity? or $\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite. — Martin Sleziak, Sep 14 '19 at 13:42

score 0 · Answer 1 · answered Sep 12 '19 at 16:45

0

If $n\ge 3$ there are more than $n/2$ integers from $1$ to $n$ that are greater than $n/3.$

So $n\ge 3 \implies n!>(n/3)^{n/2}\implies n!^{1/n}>((n/3)^{n/2})^{1/n}=\sqrt {n/3}\,.$

answered Sep 12 '19 at 16:45

DanielWainfleet

57,985

The limit superior of R given as lim sup of $\frac{1}{\left| \left(\frac{1}{n!}\right)^\frac{1}{n} \right|}$ as n approaches infinity

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