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How to prove that the sequence $a_n=n^{1/n}$ is convergent using definition of convergence?

Ramana Venkata
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    Have you proven that limits respect continuous functions? Because then $$\lim_{n\rightarrow \infty} n^{\frac{1}{n}}=\lim_{n\rightarrow\infty} e^\frac{\log n}{n}=\exp\left({\lim_{n\rightarrow\infty} \frac{\log n}{n}}\right)$$ – Eric Naslund Oct 27 '11 at 12:19
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    I doubt that is directly using the definition of convergence. – Listing Oct 27 '11 at 12:20
  • No I know that proof I am trying for proof using Definition of Convergence. – Ramana Venkata Oct 27 '11 at 12:21
  • @EricNaslund: how do you prove convergence of $\frac1n \log n$ by the definition? – SBF Oct 27 '11 at 12:31
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    @Gortaur: That depends, what is your definition of $\log n$? – Eric Naslund Oct 27 '11 at 12:32
  • @EricNaslund: exactly ) do we know that using $\log$ and $\exp$ here is allowed? – SBF Oct 27 '11 at 12:35
  • Why isn't it allowed to use log and exp? If it is not, they should have stated so in the question ! –  Oct 27 '11 at 12:52
  • Related: http://math.stackexchange.com/questions/28348/proof-that-lim-n-rightarrow-infty-sqrtnn-1/28351#28351 – JavaMan Oct 27 '11 at 17:48
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    Can't you just use L'Hopital as done here?: https://www.quora.com/Whats-the-limit-of-1-n-1-n-as-n-tends-to-infinity/answer/Manas-Uniyal?ch=10&share=78ad09e8&srid=zbEpb – Joseph Garvin Dec 26 '18 at 21:33

4 Answers4

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To show $\lim\limits_{n \to \infty} n^{1/n} = 1$, we need to show that for all $\varepsilon > 0$, there exists $N \in \mathbb{N}$ s.t. $\mid n^\frac{1}{n} - 1 \mid < \varepsilon$ for any $n > N$.

Expand \begin{equation} \mid n^\frac{1}{n} - 1 \mid < \varepsilon \Leftrightarrow (1- \varepsilon)^n < n < (1+\varepsilon)^n \end{equation}

The left part holds as $(1- \varepsilon)^n < 1 \le n$ For the right part, we use the binomial theorem, $$ (1+\varepsilon)^n = \sum_{i=0}^n \binom{n}{i}\varepsilon^i > \binom{n}{2}\varepsilon^{2} = \frac{n(n-1)}{2} \varepsilon^2 $$ Set $N = \lceil \frac{2}{\varepsilon^2} + 1 \rceil$, we have $$ (1+\varepsilon)^n > \frac{n(n-1)}{2} \varepsilon^2 > n $$

Huayu Zhang
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Noticing that $n^\frac{1}{n} > 1$ for all $n$, it all comes down to showing that for any $\epsilon > 0$, there is a $n$ such that $(1+\epsilon) \geq n^\frac{1}{n}$, or by rearranging, that

$$ (1+\epsilon)^n \geq n $$

Now, let's first of all choose an $m$ such that $(1+\epsilon)^{m}$ is some number bigger than 2, let's say the smallest number greater than $3$ that you can get. From here, swap $m$ for $2m$. This will make the left side a little over 3 times larger, and the right side 2 times larger. The next doubling will still double the right side, but the left side will increase roughly 9-fold. Repeating, we can easily see that the left side will at some point overtake the right side, and we have our $n$

Arthur
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  • Where did you use the definition here? – Ramana Venkata Oct 27 '11 at 12:59
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    The definition is used in the first sentence. Really, the definition demands that I find an $n$ such that $|n^\frac{1}{n} - 1| \leq \epsilon$. But as noted, $n^\frac{1}{n} > 1$, so we can do without the absolute value signs. Now, adding 1 to each side gives my inequality. – Arthur Oct 27 '11 at 14:16
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    The definition also states that I should find an $N$ such that for all $n> N$, this holds. A small argument about monotony like in Eric Naslund's answer will take care of that. – Arthur Oct 27 '11 at 14:23
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    I think this is the right elementary proof of this proposition and the one that immidiately came to my mind, when I saw this question. +1 for writing it down. – Sam Oct 27 '11 at 15:05
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So here is an outline of a proof:

Step 1: Notice that $n^\frac{1}{n}\geq 1$ for all $n$.

Step 2: Prove that $a_n$ is monotonically decreasing for $n\geq 3$. Equivalently we need to show that $n^{(n+1)}>(n+1)^n$.

Step 3: Show that there is a subsequence which converges to $1$. I managed to do this by considering $b_n={a_{2^{2^n}}}$. (It does not appear well in LaTeX as there are too many nested exponents. I had typed this part out, but decided to remove it)

From these three facts you can conclude that the limit is $1$.

Eric Naslund
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    One possible argument to show that it is decreasing (for $n\ge 3$) is by induction. Inductive step - by contradiction: Suppose that $(n-1)^n\ge n^{n-1}$ but $(n+1)^n> n^{n+1}$. Multiplying the two inequalities we get $(n^2-1)^n > n^{2n}$, which is a contradiction. (I kind of like this argument - which is the reason I posted it here. And I posted it as a comment, since I did not want bump this thread.) – Martin Sleziak Jun 07 '12 at 09:43
  • @MartinSleziak Why is $(n+1)^n> n^{n+1}$ that doesn't seem obvious to me. There are $n+1$ factors of $n$ on the right side. – john Feb 27 '21 at 08:24
  • @john We are actually trying to prove the opposite inequality $(n+1)^n\le n^{n+1}$. But the proof is by contradiction - so we assume $(n+1)^n\gt n^{n+1}$ and we try to arrive to a contradiction. The proof suggested in the previous comment is explained in more detail in this answer. – Martin Sleziak Feb 27 '21 at 08:36
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Well, the easiest proof is that the sequence is decreasing and bounded below (by 1); thus it converges by the Monotone Convergence Theorem...

The proof from definition of convergence goes like this:

A sequence $a_{n}$ converges to a limit L in $\mathbb{R}$ if and only if $\forall \epsilon > 0 $, $\exists N\in\mathbb{N}$ such that $\left | L - a_{n} \right | < \epsilon$ for all $n \geq N $.

The proposition: $\lim_{n\to\infty} n^{1/n} = 1 $

Proof: Let $\epsilon > 0$ be given. Then by Archimedean property of the real numbers, there exists $M \in \mathbb{N}$ such that $M < \epsilon + 1$ then find $x\in\mathbb{R}; x>2$ such that $1+M>x^{1/x}$ and let $P = \left \lceil x \right \rceil$. Then, since $f(x)=x^{1/x}$ is decreasing (for $x>e$) (trivial and left to the reader :D) take any $x\in\mathbb{N}$ such that $x>P$ and observe that (because of our choice and $M$ and $P$) we have $n^{1/n} \leq P^{1/P} \leq M \le 1 + \epsilon$ whenever $n\geq P$ and so $\left | 1 - a_{n} \right | < \epsilon$ whenever $n\geq P$. Thus $a_{n}$ converges (to 1).

Edit: We can not always find a natural number M such that $M < \epsilon$ (what if $0 < \epsilon < 1$)? But we can always find a natural number M such that $M < \epsilon + 1$.

Celia Zhu
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scibuff
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