How to prove that the sequence $a_n = (1/n)^{1/n}$ converges?

Question

Using a calculator I can see that the sequence $$ a_n =\left(\frac{1}{n}\right)^{\frac{1}{n}} $$ converges to $1$ as $n$ approaches infinity.

I would like to know the correct way to prove this.

It can also be seen that it approaches $0^0$, is that enough to say that it approaches $1$? As anything to the power $0$ is $1$?

Edit: My logic regarding $0^0$ doesn't work as suggested by @Henning Makholm as the exponential function is not continuous at (0,0)

You will find much disagreement about whether $0^0$ is even defined. But even if you do define it correctly as $0^0=1$, the exponentiation operation is not continuous at $(0,0)$, and thus the fact that $0^0=1$ tells you nothing at all about your limit. — hmakholm left over Monica, Nov 12 '18 at 13:18
But is the function continuous for any n>0? If so, the limit should approach 1 right? If not, why not? — Neo, Nov 12 '18 at 13:20
Because $0\not>0$ and exponentiation is not continuous at (0,0). — hmakholm left over Monica, Nov 12 '18 at 13:21
I can agree that it is not continuous at 0 and my reasoning that about $0^0$ is incorrect, what I am asking however is: does it even converge? If so, does it converge to 1, regardless of what $0^0$ is? — Neo, Nov 12 '18 at 13:26

score 4 · Accepted Answer · answered Nov 12 '18 at 13:18

4

$$ \lim_{n\rightarrow\infty}e^{\frac{1}{n}\ln\frac{1}{n}}=\lim_{n\rightarrow\infty}e^{-\frac{\ln n}{n}}=e^0=1. $$

answered Nov 12 '18 at 13:18

Jon

5,684

score 2 · Answer 2 · answered Nov 12 '18 at 13:24

Note that $$ \left(\frac{1}{n}\right)^{1/n}=\frac{1}{\sqrt[n]{n}}. $$ All you need to show is that $$ \lim_{n\to\infty} \sqrt[n]{n}=1. $$

Now, take a look at this question and its various answers:

Proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$

score 1 · Answer 3 · answered Nov 12 '18 at 13:28

Also $$ \left(\frac{1}{n}\right)^{\tfrac{1}{\log n}} $$ has the form $0^0$, but $$ \log\left(\left(\frac{1}{n}\right)^{\tfrac{1}{\log n}}\right)= \frac{1}{\log n}\log\frac{1}{n}=-1 $$ so the sequence converges to $e^{-1}$

In your case $$ \log\left(\left(\frac{1}{n}\right)^{\tfrac{1}{n}}\right)= -\frac{\log n}{n} $$ converges to $0$, so your sequence converges to $e^0=1$.

How to prove that the sequence $a_n = (1/n)^{1/n}$ converges?

3 Answers3