• HF08
In summary, the problem is to show that the series \sum_{k}x^{k} has a positive radius of convergence given that {a_{k}}^{\infty}_{k=0} is a bounded sequence of real numbers. The ratio test cannot be used since the terms are not nonnegative, but the root test can be used. The theorem related to the root test states that R = lim sup [1/|a_k|^1/k ]. By showing that the sequence is bounded, we can conclude that the radius of convergence is positive. Additionally, it is enough to show that the series converges for some non-zero x to prove that the radius of convergence is positive.
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

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!

morphism said:
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!

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

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

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]?

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

## What is the positive radius of convergence?

The positive radius of convergence, denoted as Rpos, is a measure of how far away from the center of a power series a function can be evaluated while still maintaining convergence.

## How is the positive radius of convergence calculated?

The positive radius of convergence is typically found by using the Ratio Test, which compares the absolute value of adjacent terms in a power series to determine if the series converges or diverges. The resulting limit will give the value of Rpos.

## What does a positive radius of convergence indicate?

A positive radius of convergence indicates that the power series will converge within a certain distance from the center. This means that the function represented by the power series will have a well-defined, finite value at all points within that radius.

## Can the positive radius of convergence change for different functions?

Yes, the positive radius of convergence can vary for different functions. It depends on the coefficients and powers of the terms in the power series. Some functions may have a larger or smaller positive radius of convergence than others.

## What happens if the positive radius of convergence is equal to zero?

If the positive radius of convergence is equal to zero, it means that the power series will only converge at the center point and nowhere else. This typically occurs when the terms in the series increase too quickly and the series diverges for all other values.

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