1. Feb 5, 2008

### HF08

Problem:

Suppose that {$$a_{k}$$}$$^{\infty}_{k=0}$$ is a bounded sequence
of real numbers. Show that $$\sum$$a$$_{k}$$x$$^{k}$$ has a

Work:

I have attempted to use the ratio test and failed. I am suspicious I can try the root
test, but I am not sure how to work it. I just got used to 'Math Type Lite' and I am
not used to Latex, hence it took me a while to type it up. Pardon if my question looks
weird.

Anyway, I am depressed. I spent the past two hours on this problem and I am getting nowhere.

Thank You,
HF08

2. Feb 5, 2008

### morphism

Look at the hypotheses of the ratio test: you can use it only when you have a series whose terms are nonnegative. This could serve as a hint!

3. Feb 5, 2008

### HF08

I am thinking of using the root test. I know there is a theorem related to the root test that discusses R = lim sup [1/|a_k|^1/k ]. (Pardon my crappy lack of knowledge of Latex, I'm learning).

So, what am I trying to say? I am not sure. I know that if a sequence is bounded, then
the sequence should be less than some constant M, always. Yet, so what? How does
this tell me that my R (radius of convergence) is positive?

I hope this tells you where I have failed. Ugh, I really hate power series.

HF08

4. Feb 5, 2008

### morphism

Suppose $|a_k| <= M$ for all k, where M>0. Then $|a_k|^{1/k} \leq M^{1/k}$ and thus $\limsup_k |a_k|^{1/k} \leq \sup_k M^{1/k} < \infty$.

5. Feb 5, 2008

### HallsofIvy

Staff Emeritus
Notice that all you really need to do is show that the series converges for some non-zero x. Then the radius of convergence must be larger than |x| and so is positive.

You are told that the sequence {an} is bounded. That is, there exist some positive integer M such that |an|< M. What is the radius of convergence of
[tex]\sum_{n=0}^{\infty} Mx^n[/itex]?

6. Feb 5, 2008

### HF08

Thanks

Thank you all for your help. I think I got a solid result. I shored up the arguments used here with the root test. :-)

HF08