Show that $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$.

I have to show this using the AM/GM inequality, imitating the proof that $\lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1$ for $x>0$ and considering $n^{\frac{1}{2n}}$.

This is how I proved that $\lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1$ for $x>0$:

The Arithmetic/Geometric mean inequality states that, given non-negative $a_{1}, a_{2}, a_{3}, ... a_{n}$,

$GM = (a_{1} a_{2} ... a_{n})^\frac{1}{n} \leq \frac{a_{1} + a_{2} + ... + a_{n}}{n} = AM$.

Now I apply the AM/GM inequality to the numbers $a_{1} = a_{2} = ... = a_{n-1} = 1$ and $a_{n} = x$.

Then $G_{n} = x^{\frac{1}{n}}$ and $A_{n} = \frac{(n - 1 + x)}{n} = \frac{(x - 1)}{n} + 1$.

Hence $x^\frac{1}{n} \leq \frac{x - 1}{n} + 1$

i.e. $x^\frac{1}{n} - 1 \leq \frac{x - 1}{n}$.

Now, if $x \geqslant 1$, then

$0 \leq x^\frac{1}{n} -1 \leq \frac{x - 1}{n}$

and hence $x^\frac{1}{n} \rightarrow 1$ as $n \rightarrow \infty$ by the sandwich theorem.

If $0 < x < 1$, then $x = y^{-1}$ where $y > 1$. But $y^\frac{1}{n} \rightarrow 1$ as $n \rightarrow \infty$ and hence

$x^\frac{1}{n} = \frac{1}{y^\frac{1}{n}} \rightarrow \frac{1}{1} = 1$ as $n \rightarrow \infty$ by the combination theorem.

Any ideas how to prove $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$ imitating the above proof and considering $n^{\frac{1}{2n}}$? I'd appreciate any help!

2. Originally Posted by clubbed2death

Show that $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$.

I have to show this using the AM/GM inequality, imitating the proof that $\lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1$ for $x>0$ and considering $n^{\frac{1}{2n}}$.

This is how I proved that $\lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1$ for $x>0$:

The Arithmetic/Geometric mean inequality states that, given non-negative $a_{1}, a_{2}, a_{3}, ... a_{n}$,

$GM = (a_{1} a_{2} ... a_{n})^\frac{1}{n} \leq \frac{a_{1} + a_{2} + ... + a_{n}}{n} = AM$.

Now I apply the AM/GM inequality to the numbers $a_{1} = a_{2} = ... = a_{n-1} = 1$ and $a_{n} = x$.

Then $G_{n} = x^{\frac{1}{n}}$ and $A_{n} = \frac{(n - 1 + x)}{n} = \frac{(x - 1)}{n} + 1$.

Hence $x^\frac{1}{n} \leq \frac{x - 1}{n} + 1$

i.e. $x^\frac{1}{n} - 1 \leq \frac{x - 1}{n}$.

Now, if $x \geqslant 1$, then

$0 \leq x^\frac{1}{n} -1 \leq \frac{x - 1}{n}$

and hence $x^\frac{1}{n} \rightarrow 1$ as $n \rightarrow \infty$ by the sandwich theorem.

If $0 < x < 1$, then $x = y^{-1}$ where $y > 1$. But $y^\frac{1}{n} \rightarrow 1$ as $n \rightarrow \infty$ and hence

$x^\frac{1}{n} = \frac{1}{y^\frac{1}{n}} \rightarrow \frac{1}{1} = 1$ as $n \rightarrow \infty$ by the combination theorem.

Any ideas how to prove $\lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$ imitating the above proof and considering $n^{\frac{1}{2n}}$? I'd appreciate any help!
let $n \in \mathbb{N}.$ put $a_1 = \cdots = a_{n-1}=1, \ a_n = \sqrt{n}.$ then by AM-GM: $(\sqrt{n})^{\frac{1}{n}} \leq \frac{n -1+\sqrt{n}}{n}=1 + \frac{\sqrt{n} - 1}{n}.$ thus: $n^{\frac{1}{n}} \leq \left(1 + \frac{\sqrt{n}-1}{n} \right)^2.$

hence: $1 \leq n^{\frac{1}{n}} \leq \left(1 + \frac{\sqrt{n}-1}{n} \right)^2.$ now apply the squeeze theorem to finish the proof.

3. Look here.

4. Cheers NonCommAlg ! Can't believe I didn't think of using $a_{n} = \sqrt{n}$ to solve it.