prove that $\sqrt[n]{(n)}$ converges

Question

Prove that the sequence $a_n = n^{\frac{1}{n}}$ converges.

My plan was to prove that this sequence is non-increasing and limited, therefore the limit would be the $limInf(a_n)$. To do so, the best I could think was:

for every $n>e$, $n>(1+\frac{1}{n})^n$ because the sequence $(1+\frac{1}{n})^n$ is increasing, so if n is higher than the limit, it's higher than every term.

$$n>\left(1+\frac{1}{n}\right)^n \Rightarrow n^{1/n}>\left(\frac{n+1}{n}\right) \Rightarrow n^{1+\frac{1}{n}}>(n+1) \Rightarrow n^{\frac{n+1}{n}}>(n+1)$$ $$\therefore n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$$

and since $n>0 \Rightarrow n^{\frac{1}{n}}>0$ my sequence is non crescent and limited, therefore it's convergent.

My problem actualy is that, to use this, I must first prove that $(1+\frac{1}{n})^n$ is crescent, which I couldn't manage to do. Can anyone find a easier way to prove this?(or show that $(1+1/n)^n$ is increasing?)

Answer 1

Actually $\lim\limits_{n \to \infty} n^{1/n} = 1$.

Let us prove this using the definition of a limit: $$\forall \epsilon > 0, \exists n_0 \in \mathbb{N} \text{ such that } \vert n^\frac{1}{n} - 1 \vert < \epsilon, \quad \forall n > n_0$$

Notice that the following are equivalent

(a) $\mid n^\frac{1}{n} - 1 \mid < \epsilon $

(b) $(1- \epsilon)^n < n < (1+\epsilon)^n$

Note that $(1- \epsilon)^n < 1 \le n$. As for $(1 + \epsilon)^n$, use Binomial theorem as follows $$ (1+\epsilon)^n = \sum_{i=0}^n \binom{n}{i}\epsilon^i > \binom{n}{2}\epsilon^{2} = \frac{n(n-1)}{2} \epsilon^2 $$ Fix $n_0 = \text{ceil} \big(1 + \frac{2}{\epsilon^2} \big)$, we have $$ n < \frac{n(n-1)}{2} \epsilon^2 <(1+\epsilon)^n $$

Answer 2

Hint. Note that $x_n=n^{1/n}$ can be written as $1+p_n=n^{1/n}$. How can the Sandwich theorem be used here?

Answer 3

Your idea is good and you only need to prove that $(1 + 1/n)^{n}\leq 3$ which is easy if we use binomial theorem. We have \begin{align} a_{n} &= \left(1 + \frac{1}{n}\right)^{n}\notag\\ &= 1 + 1 + \frac{n(n - 1)}{2! n^{2}} + \cdots\notag\\ &= 2 + \frac{1 - 1/n}{2!} + \frac{(1 - 1/n)(1 - 2/n)}{3!} + \cdots\notag\\ &\leq 2 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!}\notag\\ &\leq 2 + \frac{1}{2} + \frac{1}{2^{2}} + \cdots + \frac{1}{2^{n - 1}}\notag\\ &= 2 + \frac{1/2 - (1/2)^{n}}{1 - (1/2)}\notag\\ &< 2 + \frac{1/2}{1 - 1/2}\notag\\ &= 3\notag \end{align} Hence if $n \geq 3$ then $$\left(1 + \frac{1}{n}\right)^{n} \leq 3 \leq n$$ and this is equivalent to $$n^{1/n} \leq (n + 1)^{1/(n + 1)}$$ So the sequence $s_{n} = n^{1/n}$ is non-increasing and obviously we have $s_{n} \geq 1$ and hence $\lim_{n \to \infty}s_{n}$ exists (i.e. $s_{n}$ converges to some limit $L$ and $L \geq 1$). This answers your question.

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But we can continue further and it is easy to prove that $L = 1$ because $L > 1$ would lead to a contradiction. Since $s_{n}$ is non-increasing it follows that $s_{n} = n^{1/n} \geq L$ or $n \geq L^{n}$ for all $n \geq 3$. If $L > 1$ then we know that $L^{n}/n^{k} \to \infty$ for all $k > 0$ and hence putting $k = 1$ we see that we can't have $L^{n} \leq n$.