# Showing a sequence is monotonically increasing

1. Oct 30, 2013

### bonfire09

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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 im 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 im stuck on.

2. Oct 30, 2013

### Dick

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?

3. Oct 30, 2013

### bonfire09

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 im not sure from here.

Last edited: Oct 30, 2013
4. Oct 30, 2013

### Dick

It's $\frac{c^{1/(k+1)}}{c^{1/k}}$. Use the laws of exponents and combine them.

5. Oct 30, 2013

### bonfire09

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.