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Show that $\lim_{n\to \infty} n^{1/n} = 1$.

My attempt

Let $a_{n} = n^{1/n}$.

$|a_{n}-1| = |n^{1/n}-1| < n^{1/n} \leq n$.

Consider $|a_{n}-1| < \varepsilon$.

or $n<\varepsilon$.

or $n > 1/\varepsilon$.

Let $m$ be any integer greater than $1/\varepsilon$. Then for $\varepsilon>0$, there exists a positive integer $m$ such that $|a_{n}-1| < \varepsilon$, for all $n\geq m$.

Therefore, $\lim_{n\to \infty} n^{1/n} = 1$.

Is this proof correct?

Shaun
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Sasikuttan
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    No... you want $n$ large and yet you are supposing $n\lt\epsilon$ for arbitrarily small $\epsilon$. This does not work – FShrike Jan 24 '22 at 17:49
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    Not even close. $n<\epsilon$ is not equivalent to $n>\frac{1}{\epsilon}$, but to $\frac{1}{n}>\frac{1}{\epsilon}$. Anyway, if $\epsilon<1$ then $n<\epsilon$ clearly never happens. – Mark Jan 24 '22 at 17:49
  • It doesn't look right...at all. How can you prove that $;m>\frac1\epsilon\implies |a_n-1|<\epsilon;$ ?How deos this follow from what you did? – DonAntonio Jan 24 '22 at 17:49
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    No. First, you want to show $|a_n-1|<\epsilon,$ not assume it. But also, $a<b$ and $a<c$ tells you nothing about whether $b=c,$ $b<c$ or $c<b,$ so it is unclear how you get $n<\epsilon.$ – Thomas Andrews Jan 24 '22 at 17:50
  • The only property of $n^{1/n}$ that you have used was that $n^{1/n}\leqslant n$. So, if your proof was correct, it would prove that $\lim_{n\to\infty}n=1$. Do you really believe in that? – José Carlos Santos Jan 24 '22 at 17:51
  • @Mark Oops! I forgot to take the reciprocal of 'n'. My bad! – Sasikuttan Jan 24 '22 at 17:52
  • But why would someone downvote this? The whole point of asking this is to find out where I went wrong. – Sasikuttan Jan 24 '22 at 17:55
  • @Curiouserandcuriouser In fairness you did provide your working. I expect the downvoter skim-read your post and judged it too quickly as a "low-quality post", i.e. they felt you hadn't provided much motivation or working, as many such posts appear at a glance similar to your own – FShrike Jan 24 '22 at 18:00
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    "$n < \epsilon$" should be your tip off that it couldn't possibly work. $n$ gets arbitrarily large but we need $\epsilon$ to be arbitrarily small.... But so far as I can tell the only reason you would think that $n < \epsilon$ is because $|a_n-1|<\begin{cases}\epsilon\ n\end{cases}$ and you conclude $n < \epsilon$. But $M<\begin{cases}U\ S\end{cases}$ doesn't tell us anything about how $U$ and $S$ relate. – fleablood Jan 24 '22 at 18:16
  • It's true that $|n^{\frac 1n} - 1|< n$ but as $n$ gets large rather than small the gets you nowhere. It's a bit like saying $|7 - \frac 1n| < 5$ million (so $\lim \frac 1n= 7$????) Overly large comparisons get you no-where.... – fleablood Jan 24 '22 at 18:29
  • To show this limit by the very definition can prove to be pretty hard. There are other, more promising and perhaps easier, ways to prove it. – DonAntonio Jan 24 '22 at 18:40
  • It's important to realize the difference between going forward: $n > N\implies ..... \implies |a_n -1| < \epsilon$ and going backwards $|a_n-1|< \epsilon \Leftarrow .... \Leftarrow n> M$. You use the second way to figure out how to figure out $N$ in terms of $\epsilon$. But the big difference is going forward the $\implies$ are all "from this we can always conclude that". Going backwards is the exact opposite: "to conclude that, one of the ways we can do it is this".... to be continued.... – fleablood Jan 24 '22 at 19:24
  • ... cont...For instance "Billy is a lion $\implies$ Billy eats meat". Whenever Billy is a lion then Billy has to eat meat. "Billy eats meat $\Leftarrow$ Billy is a lion" means we want to conclude Billy eats meat; how can we do that; well, one way would be if Billy is a lion; Billy eats meat is implied *BY* Billy being a lion. So if we want $|a_n-1|< \epsilon \Leftarrow .... \Leftarrow n> M$ we are saying we want to conclude $|a_n-1|$ and we can get there from $n> N$. ... to be continued .... – fleablood Jan 24 '22 at 19:29
  • ... continued.... the is important with inequalities. We can have $1 < x < 3\implies 0 < x < 5$ because if $1 < x$ and $0< 1$ then $0< x$ and same with $x<3$ and $3<5$ implies $0<x<5$. But going backwards we'd have $0< x < 5 \Leftarrow 1< x <3$ because to start with our conclusion $0<x<5$ is implied by $1<x<3$. SO that is one way to get there. So an $N,\epsilon$ proof backwards might go like this $|a_n-1|<\epsilon\Leftarrow 1-\epsilon<a_n<1+\epsilon\Leftarrow <1-\epsilon<somethingtodowithN<a_n<somethingtodowithN<1+\epsilon\Leftarrow n>N$. – fleablood Jan 24 '22 at 19:38
  • I assume you are trying to prove that, given $\ \varepsilon >0\ $ there exists $\ n\in\mathbb{N}\ $ such that $\ \left \vert n^{\frac{1}{n}}-1 \right\vert < \varepsilon.$ Unfortunately, I don't even see the idea behind your proof attempt, let alone a logical sequence of steps. – Adam Rubinson Jan 24 '22 at 18:47
  • @fleablood Thank you. I'm kind of new to Real Analysis. Although I do have some fair idea about how this works, the fact that this proof using epsilon is a bit "backwards" seems to be tripping me up. I think I now have a better understanding of the sequence of steps followed. – Sasikuttan Jan 24 '22 at 19:45
  • The simplest proof: Let $a_n=1+b_n. $ Then when $n\ge 2$ we have $b_n>0,$ and by the Binomial Theorem $n=(1+b_n)^n=$ $\sum_{j=0}^n\binom n j b_n^j>$ $ \binom n 0+\binom n 2 b_n^2=$ $1+\frac {n(n-1)}{ 2} b_n^2,$ implying $\frac 2 n> b_n^2.$ I gave you a +1 for effort even though your answer makes no sense. – DanielWainfleet Jan 24 '22 at 19:53
  • @DanielWainfleet But when you say $a_{n} = 1 + b_{n}$, aren't you assuming that $a_{n} \geq 1$, for all n belonging to N? – Sasikuttan Jan 25 '22 at 03:34
  • If $0\le x\le 1$ and $n\in\Bbb N$ then $x^n\le 1.$ So if $n\ge 2$ and $0\le x=a_n=1+b_n=n^{1/n}$ then $x>1,$ otherwise we would have $1\ge x^n=n.$ – DanielWainfleet Jan 25 '22 at 20:55

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