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alright so im supposed to show that the limit of $\displaystyle \lim_{n \to \infty} n^{\frac{1}{n}} = 1$ , ive alwasy had problems with math questions of this type: "show" and never really knew what to do, but since its $n^{\frac{1}{n}}$ thats basically $n$th-root of $n$ when $\lim \to \infty =1$ of that function so yeah :)

I have no idea how to go forward with this problem, but I managed to figure out that if $n=2$ then I get $1.4$, if $n=5$ then I get $1.38$ and if $n=1000$ then I get $1.007$ so it seems like im getting closer and closer to $1$ as $x$ approach larger numbers?? ok anyway! thanks for tips and help :D

mle
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gger123
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  • @FH93: One suggestion, please try to pick a question which itself isn't closed as a dupe. Perhaps it should be a feature of this site... – Aryabhata Mar 12 '14 at 01:44
  • @Aryabhata That would be sensible in future! Thanks for the heads up. – FH93 Mar 12 '14 at 01:46

5 Answers5

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Hint: Note that $n^{1/n} \geq 1$ for $n\geq 1$ so there is a sequence of non-negative reals $\delta(n)$ s.t. $n^{1/n} = 1 + \delta (n)$ for each $n$. Show that $\delta(n) \to 0$ as $n\to \infty$ by raising both sides to the power of $n$.

Solution: In particular, note that:

$n = (1+\delta (n))^n = 1+n\delta (n) + \dfrac{n(n-1)}{2}[\delta(n)]^2 + … + [\delta (n)]^n$

$\implies n \geq \dfrac{n(n-1)}{2}[\delta(n)]^2$ (As all other terms are non-negative)

$\implies 0\leq \delta (n) \leq \sqrt{\dfrac{2}{n-1}} \to 0$ as $n\to \infty$

$\implies \delta (n) \to 0$ as $n\to \infty$ (By squeezing)

$\implies 1+ \delta (n) \to 1$ as $n \to \infty$

Hence $\displaystyle\lim_{n\to \infty} n^{1/n} = 1$, as desired.

FH93
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  • hmm what? do you mean I should just raise n^(1/n)^n = 1^n`? – gger123 Mar 12 '14 at 00:53
  • @gger123 Not quite. I mean that you should note $n^{1/n} = 1+\delta (n) \implies n = (1+\delta (n))^n$. – FH93 Mar 12 '14 at 00:55
  • alright, but Im curious about the theory here, how did you find n^(1/n)=1+δ(n)? :D – gger123 Mar 12 '14 at 00:58
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    @gger123 As explained in the first line, we know that $n\geq 1 \implies n^{1/n} \geq 1^{1/n} = 1$ i.e. $n^{1/n}$ is either equal to 1 or strictly larger than it. This can be summarised by noting that then, for each $n$, there exists a non-negative real number (that we call $\delta (n)$) with the property that $n^{1/n} = 1 + \delta (n)$.

    These $\delta (n)$ naturally form a sequence as $n$ increases over the naturals; and by considering the behaviour of this sequence as $n$ gets large, we can also determine the behaviour of $n^{1/n}$.

     – FH93 Mar 12 '14 at 01:06
  • It's also a good idea to get a feel for the limit first - see if you can guess it and then try to make it work. $1$ is the reasonable guess that motivates this approach. – FH93 Mar 12 '14 at 01:08
  • mmh ok but, how can s(n) go to 0? for lim n->infinity? – gger123 Mar 12 '14 at 01:09
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    @gger123 Please read the post! Start off by raising both sides to the power of $n$ as explained twice already and see if you can bound $\delta (n)$ above by something that tends to $0$. – FH93 Mar 12 '14 at 01:17
  • well ok I raised both and what now? I mean only way that can be 0 is if n=0? idk what im even trying to do, your explanation is not very clear :( oops even then it doesnt get 0 since the n above the parentesis and I get 0^0=1 :O – gger123 Mar 12 '14 at 01:21
  • Well, the answer right like that with no explanation is not gonna help much, but thx I guess :d – gger123 Mar 12 '14 at 01:29
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    We're trying to show that $\delta(n)$ is small for big $n$ (since this would mean that $1+\delta(n)$ is close to 1 for big $n$, which is precisely what we mean by $n^{1/n}\to 1$ given the set up of the problem).

    One way to do this is to bound $\delta (n)$ above by some other function of $n$ that eventually becomes small itself for big $n$. We don't know much about $\delta (n)$ other than: the fact that it satisfies the equation in comment 2; and the fact that it's non-negative. Raising both sides of that equation gives us more to work with so we try that and hope this helps bound delta n.

     – FH93 Mar 12 '14 at 01:35
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    @gger123 Hopefully that clarifies things a little more. There's more than enough up there for you to solve the problem so I suggest you read over it a few times before asking for further explanation. A lot of these comments have been nothing more than repetition of what I had said previously. – FH93 Mar 12 '14 at 01:38
  • well it was more the actual math I was wondering, how u did it etc, specially the first two lines with all that complicated stuff :P like no way I know what [δ(n)]2+…+[δ(n)]n even means, like the example below is much easier for me, even theres some properties I dont understand :P – gger123 Mar 12 '14 at 01:43
  • Note that when you see a duplicate question, especially one that you answered elsewhere the proper course of action is to flag the question as a duplicate. As it is, your answer came up in the "review cue" because it was automatically flagged as a duplicate answer (reference the link above for the copy-and-paste source). When flagged as such, users conducting reviews need to decide the course of action. Since you are relatively new to answering, I'm sure all will be well. – amWhy Mar 12 '14 at 13:58
  • @amWhy Thanks for letting me know. – FH93 Mar 12 '14 at 14:29
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    No need to worry, the "auto" flag has been "invalidated" by users. And any way, the worst thing that could have happened would have been deletion of the answer. Nothing more. In all sincerity, keep up the good work! – amWhy Mar 12 '14 at 14:52
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By AM-GM and Sandwich Theorem

$\frac{1 + 1 + 1 + \dots + \sqrt{n} + \sqrt{n}}{n} \geq \sqrt[n]{n} \geq 1$

$\lim \limits_{n \to \infty} \frac{n-2 + 2 \sqrt{n}}{n}$

$\lim \limits_{n \to \infty} 1 - \frac2n + \frac{2}{\sqrt{n}}$

$\therefore \lim \limits_{n \to \infty} \sqrt[n]{n} = 1$

MT_
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  • AM-GM? :P hmm and by sandwich theorem I guess you mean squeeze theorem? – gger123 Mar 12 '14 at 01:48
  • I like this a lot. – FH93 Mar 12 '14 at 01:48
  • AM-GM states that the arithmetic mean (average) of a set of positive real numbers is greater than or equal to the geometric mean. Formally, it says $\frac{x_1 + x_2 + \dots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \dots x_n}$. – MT_ Mar 12 '14 at 01:49
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Hint: take Log and use L'hospitale rule.

DeepSea
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$$\mbox{To calculate: }\lim_{n\to\infty}n^{1/n}$$ Let $x=n^{1/n}$. Then, $\ln(x)=\dfrac{\ln(n)}{n}$. So, $$\lim_{n\to\infty}n^{1/n}=\exp\left(\lim_{n\to\infty}\dfrac{\ln(n)}{n}\right)\\ \exp\left(\lim_{n\to\infty}\dfrac{\ln(n)}{n}\right)=\exp\left(\lim_{n\to\infty}\dfrac{1}{n}\right)\mbox{(by L'Hopital's rule.) This you can show goes to $1$.}$$

  • can you explain the first part how x=n^(1/n is the same as lnx=ln(n)/n ?

    and also in the last expression right before you write by L'hopitals' rule, u sure if it should be an exp infront? doesnt that cancel out completely in order to remove the ln on the top?

     – gger123 Mar 12 '14 at 00:57
  • I cant use l'hopital unless I get 0/0?? and this give me 1/infinity – gger123 Mar 12 '14 at 01:06
  • No, that's not exactly true - unless you get an indeterminate form, you can't use L'Hopital's. Also, $1/\infty$ can be said to be $0$, and $\exp(0)=1$. So, you've got your solution. –  Mar 12 '14 at 01:08
  • ok so then the question reminds how could you just remove the ln?? did you kinda remove it from x also so we get x=1 and not lnx=1? idk – gger123 Mar 12 '14 at 01:10
  • @gger123 That's a limit property. Search it up on Google. –  Mar 12 '14 at 01:12
  • ok, search up "can I remove ln from a limit"? or what XD – gger123 Mar 12 '14 at 01:13
  • Ahh.. now I get it, you use L'hopitals rule to go from exp(lim n->infity of ln(n)/n to exp(lim n->infity of 1/n right? – gger123 Mar 12 '14 at 02:30
  • @gger123 Exactly! –  Mar 12 '14 at 04:07
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Here is another way and one of the simplest.

Since $\liminf (n+1)/n\le \liminf n^{1/n}\le\limsup n^{1/n}\le \limsup(n+1)/n$. Then $1\le \liminf n^{1/n}\le\limsup n^{1/n}\le 1$.

Thus $\liminf n^{1/n}=\limsup n^{1/n}=1$ and so $\lim n^{1/n}=1$.

It's worth to know that for $(a_n) \subset \mathbb{R}^{>0}$, If $\lim a_{n+1}/a_n= L$ where $0\le L < \infty$, then $\lim a_n^{1/n}=L$. This last assertion is obvious from the fact that

$$\liminf a_{n+1}/a_n \le \liminf a_n^{1/n}\le \limsup a_n^{1/n}\le \limsup {a_{n+1}/a_n}$$

Jose Antonio
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