Prove sequence $a_n=n^{1/n}$ is convergent [duplicate]

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

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4 Answers

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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 $n$

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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 $$ (1+\varepsilon)^n > \frac{n(n-1)}{2} \varepsilon^2 > n $$

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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 $1$.

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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 $M < \epsilon + 1$.

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