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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.

Marios Gretsas
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6 Answers6

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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.

Community
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Marios Gretsas
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  • Maybe OP does not know that $\sqrt[n]{n}\to 1$. – Robert Z Oct 24 '19 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 – Marios Gretsas Oct 24 '19 at 20:06
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    To integrate the site better, the $n^{1/n}$ case is covered here. – Jam Oct 28 '19 at 17:00
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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 $n^{1/n} > 1$, $n^{1/n} \to 1$.

marty cohen
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  • This proof is not complete... you just showed that by having $n$ goes to infinity, $n^{1/n}$ will be smaller than 1. – Mithridates the Great Oct 24 '19 at 14:56
  • Nope. My bound is $1+(3/\sqrt{n})$. – marty cohen Oct 24 '19 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 $\lim_{n \rightarrow \infty} a_{n} = \lim_{n \rightarrow \infty} b_{n} = 1$. – Mithridates the Great Oct 24 '19 at 15:40
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    We have $1 < n^{1/n} < 1+3/\sqrt{n}$. In the limit, the "<" becomes "$\le$", so the limit is 1. – marty cohen Oct 24 '19 at 18:06
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    (+1) Nice (elementary) proof. – Robert Z Oct 24 '19 at 20:02
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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 $0<y_{n}\leq\sqrt{\frac{2}{n-1}}$. 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, $\lim_{n\rightarrow\infty}(n!)^{\frac{1}{n!}}=\lim_{n\rightarrow\infty}x_{n!}=1$.

Danny Pak-Keung Chan
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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.

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

Kishalay Sarkar
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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.

Certainly not a dog
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