# How to prove the sequence $\sqrt[n!]{n!}$ has a limit

How to prove $$\sqrt[n!]{n!}$$ converges and find it's limit.

We haven't yet covered sequences with exponents and logarithms and hence cannot use them in the proof.

I have tried finding out where to start but all solutions use exp/log.

• You haven't had exponents but you are looking at $(n!)^{1/n!}$? Oct 24, 2019 at 14:05

The sequence is the post is a subsequence of $$b_n=\sqrt[n]{n}$$

So it has the same limit as $$b_n$$,which is $$1$$

The fact that $$\lim_n\sqrt[n]{n}=1$$ can be proved without exponents.

We can put $$b_n=1+t_n$$ and prove that $$t_n \to 0$$ using the binomial theorem and simple inequalities.

• Maybe OP does not know that $\sqrt[n]{n}\to 1$. Oct 24, 2019 at 20:04
• @RobertZ.. The fact that $\sqrt[n]{n} \to 1$ can be proved without using exponentials as you can see in the other answers and he did not stated what he knows,just what we cannot use to find the limit.so we cannot always assume a priori what the O.P knows or does not know..we answer the question and wait for further information,if needed Oct 24, 2019 at 20:06
• To integrate the site better, the $n^{1/n}$ case is covered here.
– Jam
Oct 28, 2019 at 17:00

To show that $$n^{1/n} \to 1$$:

By Bernoulli's inequality, $$(1+1/\sqrt{n})^n \ge 1+\sqrt{n} \gt \sqrt{n}$$ so, raising to the $$2/n$$ power,

$$\begin{array}\\ n^{1/n} &=(\sqrt{n})^{2/n}\\ &\lt (1+1/\sqrt{n})^2\\ &=1+2/\sqrt{n}+1/n\\ &\lt 1+3/\sqrt{n}\\ \end{array}$$

so, since $$n^{1/n} > 1$$, $$n^{1/n} \to 1$$.

• This proof is not complete... you just showed that by having $n$ goes to infinity, $n^{1/n}$ will be smaller than 1. Oct 24, 2019 at 14:56
• Nope. My bound is $1+(3/\sqrt{n})$. Oct 24, 2019 at 15:30
• Yes, your bound is $1+\frac{3}{\sqrt{n}}$, but: $\lim_{n \rightarrow \infty} n^{\frac{1}{n}} < \lim_{n \rightarrow \infty} (1 + \frac{3}{\sqrt{n}})$, so: $\lim_{n \rightarrow \infty} n^{\frac{1}{n}} < 1$. You need another inequality that is always smaller than $n^{\frac{1}{n}}$ and then show that sequence also will converge to 1 if $n \rightarrow \infty$ to complete your proof. Something like this: $a_{n} < n^{\frac{1}{n}} < b_{n}$ and $\lim_{n \rightarrow \infty} a_{n} = \lim_{n \rightarrow \infty} b_{n} = 1$. Oct 24, 2019 at 15:40
• We have $1 < n^{1/n} < 1+3/\sqrt{n}$. In the limit, the "<" becomes "$\le$", so the limit is 1. Oct 24, 2019 at 18:06
• (+1) Nice (elementary) proof. Oct 24, 2019 at 20:02

Let $$y_{n}=n^{\frac{1}{n}}-1>0$$. Then $$\begin{eqnarray*} n & = & (1+y_{n})^{n}\\ & = & 1+ny_{n}+\frac{n(n-1)}{2}y_{n}^{2}+\ldots\\ & \geq & \frac{n(n-1)}{2}y_{n}^{2}. \end{eqnarray*}$$ Therefore $$0. This shows that $$y_{n}\rightarrow0$$ as $$n\rightarrow\infty$$. It follows that $$\lim_{n\rightarrow\infty}n^{\frac{1}{n}}=1$$.

Denote $$x_{n}=n^{\frac{1}{n}}$$. Note that $$\{(n!)^{\frac{1}{n!}}\}$$ is just a subsequence of $$\{x_{n}\}$$. Hence, $$\lim_{n\rightarrow\infty}(n!)^{\frac{1}{n!}}=\lim_{n\rightarrow\infty}x_{n!}=1$$.

Just prove that $$\lim_{m \rightarrow \infty} m^{1/m}= \exp[\lim_{m \rightarrow \infty} m \ln m]= \exp[\lim_{m \rightarrow \infty} \frac{\ln m}{1/m}]=e^{0}=1.$$ In the last limit we have used L'Hospital's rule.

Consider the sequence $$n^{1/n}$$ which converges to $$1$$.So any subsequence converges to $$1$$.

Here is a solution without using $$\exp()$$ and $$\log()$$ as requested.

By the ratio test, $$a_n$$ converges if $$\displaystyle\lim_{n\to\infty}\left|{a_{n+1}\over a_n}\right| <1$$. Here we have

$$\displaystyle\lim_{n\to\infty}\left| {(n+1)!^{1\over(n+1)!}\over n!^{1\over n!}}\right|$$

$$=\displaystyle\lim_{n\to\infty} \left|{(n+1)!^{1\over(n+1)!}\over n!^{n+1\over (n+1)!}}\right|= \displaystyle\lim_{n\to\infty}\left|\left( {(n+1)!\over n!^{n+1}}\right)^{1\over(n+1)!}\right|=\displaystyle\lim_{n\to\infty}\left|\left( {n+1\over n!^n}\right)^{1\over(n+1)!}\right|$$ $$=\displaystyle\lim_{n\to\infty} \left|e^{\ln{n+1\over n!^n}\over n+1}\right|=\displaystyle\lim_{n\to\infty} \left|e^{{\ln(1+n)\over (1+n)}-{n\ln n!\over n+1}}\right|$$

Now, since $$\displaystyle\lim_{x\to\infty} {\ln(1+x)\over 1+x} = 0$$ and $$\displaystyle\lim_{x\to\infty} {x\over x+1}= 1$$ our limit ends up being equal to $$\left|1\over n!\right|$$ which in fact tends to $$0^+$$, so indeed is less than $$1$$. Hence by the ratio test this series converges.