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

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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, $\lim_{n\rightarrow\infty}(n!)^{\frac{1}{n!}}=\lim_{n\rightarrow\infty}x_{n!}=1$.

Answer 4

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.

Answer 5

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

Answer 6

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.