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Math Help - Proving b^{1/n} can be made arbitrarily close to 1

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    Junior Member utopiaNow's Avatar
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    Proving b^{1/n} can be made arbitrarily close to 1

    I am trying to prove that for any  b \in \mathbb{R} such that b > 1. We have that for any n \in \mathbb{Z}^+, 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: b - 1 \geq n(b^{\frac{1}{n}} - 1). And I have this so far:
    b=(b^{\frac{1}{n}})^{n}=((b^{\frac{1}{n}}-1)+1)^{n}. Which I expand by the binomial theorem to get:
    \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: b - 1 \geq n(b^{\frac{1}{n}} - 1). However to claim they are non-negative I need to use the fact that 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.
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  2. #2
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    If 0<x\leqslant1 then x^n\leqslant1 for n=1,2,3,... (easy proof by induction). Put x=b^{1/n} and you see that b^{1/n}\leqslant1\;\Rightarrow\; b\leqslant1. The contrapositive version of that is: b>1\;\Rightarrow\; b^{1/n}>1.
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    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by Opalg View Post
    If 0<x\leqslant1 then x^n\leqslant1 for n=1,2,3,... (easy proof by induction). Put x=b^{1/n} and you see that b^{1/n}\leqslant1\;\Rightarrow\; b\leqslant1. The contrapositive version of that is: 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.
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    Quote Originally Posted by utopiaNow View Post
    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 b^{1/n}>1 (which is only one minor step in the proof). You still need to follow Rudin's routine to establish that 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.
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    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by Opalg View Post
    All I was doing in my previous comment was to indicate why b^{1/n}>1 (which is only one minor step in the proof). You still need to follow Rudin's routine to establish that 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 b^{1/n}>1. Thank you for the response and assistance.
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