Showing a sequence is monotonically increasing

bonfire09
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Homework Statement



Prove that if ##0<c<1## then llim##c^{\frac{1}{n}}=1## using the monotone convergence theorem.

Homework Equations





The Attempt at a Solution


I let ##c_n=c^{\frac{1}{n}}## and it follows since ##0<c<1 \implies 0<c^{\frac{1}{n}}<1## Thus ##c_n## is bounded above by 1. Now I want to show that ##c_n## is monotonically increasing by induction but I am not sure how to do it. So for my base case I know I need to show ##c_1<c_2## And for my inductive case I suppose that ##c_k<c_{k+1}## and show ##c_{k+1}<c_{k+2}## which is what I am stuck on.
 
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bonfire09 said:

Homework Statement



Prove that if ##0<c<1## then llim##c^{\frac{1}{n}}=1## using the monotone convergence theorem.

Homework Equations





The Attempt at a Solution


I let ##c_n=c^{\frac{1}{n}}## and it follows since ##0<c<1 \implies 0<c^{\frac{1}{n}}<1## Thus ##c_n## is bounded above by 1. Now I want to show that ##c_n## is monotonically increasing by induction but I am not sure how to do it. So for my base case I know I need to show ##c_1<c_2## And for my inductive case I suppose that ##c_k<c_{k+1}## and show ##c_{k+1}<c_{k+2}## which is what I am stuck on.

I don't think you need induction at all. Just look at ##\frac{c_{k+1}}{c_k}##. Is it greater than 1 or less than 1?
 
Well ##\frac{c_{k+1}}{c_{k}}≥ 1##. But I'm not sure how to show that. I know that ##c_k=c^\frac{1}{k}\implies \frac{1}{c_k}=\frac{1}{c^\frac{1}{n}}≥1##. Then I am not sure from here.
 
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bonfire09 said:
Well ##\frac{c_{k+1}}{c_{k}}≥ 1##. But I'm not sure how to show that. I know that ##c_k=c^\frac{1}{k}\implies \frac{1}{c_k}=\frac{1}{c^\frac{1}{n}}≥1##. Then I am not sure from here.

It's ##\frac{c^{1/(k+1)}}{c^{1/k}}##. Use the laws of exponents and combine them.
 
I see ##\frac{c_{k+1}}{c_{k}}=\frac{c^\frac{1}{k+1}}{c^\frac{1}{k}}=c^{\frac{1}{k+1}-\frac{1}{k}}=c^\frac{-1}{k(k+1)}## But since ##c<1## it follows ##\frac{1}{c^\frac{1}{k(k+1)}}>1##. Thanks I got it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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