1. ## Proving b^{1/n} can be made arbitrarily close to 1

I am trying to prove that for any $\displaystyle b \in \mathbb{R}$ such that $\displaystyle b > 1$. We have that for any $\displaystyle n \in \mathbb{Z}^+$, $\displaystyle b^{\frac{1}{n}}$ can be made arbitrarily close to 1.

I'm completing the routine outlined in Rudin's book. So I'm at the step where I need to prove: $\displaystyle b - 1 \geq n(b^{\frac{1}{n}} - 1)$. And I have this so far:
$\displaystyle b=(b^{\frac{1}{n}})^{n}=((b^{\frac{1}{n}}-1)+1)^{n}$. Which I expand by the binomial theorem to get:
$\displaystyle \sum_{i=0}^{n}{n \choose k}(b^{\frac{1}{n}}-1)^{n-k}1^{k}={n \choose n}(b^{\frac{1}{n}}-1)^{n}+{n \choose 1}(b^{\frac{1}{n}}-1)^{n-1}+\cdots+{n \choose n-1}(b^{\frac{1}{n}}-1)+1.$

Where my plan was to drop all but the last two terms by claiming that they are non-negative and therefore:$\displaystyle b - 1 \geq n(b^{\frac{1}{n}} - 1)$. However to claim they are non-negative I need to use the fact that $\displaystyle b^{\frac{1}{n}} > 1$. Which is the fact I'm trying to prove.

So can anyone see what I'm doing incorrectly at this point? Thanks in advance.

2. If $\displaystyle 0<x\leqslant1$ then $\displaystyle x^n\leqslant1$ for n=1,2,3,... (easy proof by induction). Put $\displaystyle x=b^{1/n}$ and you see that $\displaystyle b^{1/n}\leqslant1\;\Rightarrow\; b\leqslant1$. The contrapositive version of that is: $\displaystyle b>1\;\Rightarrow\; b^{1/n}>1$.

3. Originally Posted by Opalg
If $\displaystyle 0<x\leqslant1$ then $\displaystyle x^n\leqslant1$ for n=1,2,3,... (easy proof by induction). Put $\displaystyle x=b^{1/n}$ and you see that $\displaystyle b^{1/n}\leqslant1\;\Rightarrow\; b\leqslant1$. The contrapositive version of that is: $\displaystyle b>1\;\Rightarrow\; b^{1/n}>1$.
Thanks for the reply, and I do realize that's a much nicer way to prove what I am trying to prove. I'm just wondering about the routine outlined in Rudin. I'm just having that 1 minor problem at that step, from that point onwards the proof flows logically. However, I don't know how to prove that 1 step mentioned above. If you could point out what I need to do differently, or see differently at that step, it would be greatly appreciated.

Thank you again for the reply.

4. Originally Posted by utopiaNow
Thanks for the reply, and I do realize that's a much nicer way to prove what I am trying to prove. I'm just wondering about the routine outlined in Rudin. I'm just having that 1 minor problem at that step, from that point onwards the proof flows logically. However, I don't know how to prove that 1 step mentioned above. If you could point out what I need to do differently, or see differently at that step, it would be greatly appreciated.
All I was doing in my previous comment was to indicate why $\displaystyle b^{1/n}>1$ (which is only one minor step in the proof). You still need to follow Rudin's routine to establish that $\displaystyle b^{1/n}$ actually converges to 1. So my comment was intended to show "how to prove that 1 step mentioned above". It doesn't do more than that.

5. Originally Posted by Opalg
All I was doing in my previous comment was to indicate why $\displaystyle b^{1/n}>1$ (which is only one minor step in the proof). You still need to follow Rudin's routine to establish that $\displaystyle b^{1/n}$ actually converges to 1. So my comment was intended to show "how to prove that 1 step mentioned above". It doesn't do more than that.
Oh I see, thank you. I misunderstood the statement you gave. I, for some reason read it as a quick proof for convergence to 1, when clearly all you showed that was $\displaystyle b^{1/n}>1$. Thank you for the response and assistance.