Real analysis:limit superior problem

[K29]

Homework Statement

Let c>1 and let c_{n}=\sqrt[n]{c}-1
Show that c_{n} \geq 0 and that
\stackrel{limsup}{_{n \rightarrow \infty}}c_{n} \leq 0 by using Bernoullis inequality

(This problem actually occurs in a section on power series and there are more questions that follow from this one)

Homework Equations

(1+x)^{n} \geq 1+nx \forall x\geq -1

The Attempt at a Solution

I could do the first part, but I'm really at a loss for the second part, even with the bernoullis inequality hint.
I mean I have a fairly good understanding of limit superior so I don't even see how the result they get is even possible. In my head the limsup here is the same as the limit, and that should be greater than 0 anyway.

Either way, can you see a way to write \sqrt[n]{c}-1 so that I can use bernoullis inequality? All I can think of is pulling out -1, and that doesn't achieve much.

Any help on any of the above issues would be appreciated

 

[micromass]

Hi K29!

The thing you write down, i.e. c_n\geq 0 and \limsup{c_n}\leq 0 is of course only possible if the limsup equals zero! So what the question actually asks you to prove is that \limsup{c_n}=0.

Now, what if you apply Bernouilli's inequality to x=\sqrt[n]{c}-1?

 

[hunt_mat]

You know that c_{n}=c^{\frac{1}{n}}-1 The a little re-arrangement shows that c=(1+c_{n})^{n}, this is now in a good position to use Bernoullie inequality.

 

[K29]

Thanks Micromass for helping me understand what my goal was. (It did seem rather impossible for the limsup to be <0 and that threw me off.)
And thanks hunt_mat, that bernoulli idea helped me solve the problem and get the limsup(c_n)=0