# Behaviour of series (radius of convergence)

1. Apr 26, 2014

### Lengalicious

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

Series:
$$\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{(x)^n}{na^n}$$

what is the behaviour of the series at radius of convergence $$\rho_o=-z$$ ?

2. Relevant equations

3. The attempt at a solution
So I can specify that the series is monatonic if z is non negative as $$\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{(-z)^n}{na^n}$$ right?

But then I suppose I have to do the integral test but im a bit confused because you cannot integrate $$(-1)^{(x+1)}\frac{(-z)^x}{xa^x}$$?

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Last edited: Apr 26, 2014
2. Apr 26, 2014

### vela

Staff Emeritus
It's not clear to me what the various symbols mean. What is $\rho_0$? What is $z$? How are they related to the series and $x$?

Most of the tests you have for convergence apply to positive or non-negative series. You don't have that here, do you? Or do you?

3. Apr 26, 2014

### Dick

I think you should figure out what the radius of convergence $\rho_0$ of the given series is first. The ratio test should work nicely. Then substitute and simplify.

4. Apr 26, 2014

### Lengalicious

Sorry for being unclear, I found the radius of convergence to be $$\rho_o=|a|$$ using the ratio test, forgot to include that in the opening post. I am unclear as to how I am supposed to figure the behaviour based on this information, the question literally gets me to find the radius of convergence from the series I mentioned and figure the behaviour of the series when $$z=-\rho_o$$. I have added an image attachment of the question to clear things up hopefully.

Last edited: Apr 26, 2014
5. Apr 26, 2014

### Dick

I THINK they want you to substitute -|a| for x in the series and determine convergence of the resulting series. The result will depend on whether you take 'a' to positive or negative. But the series will simplify a lot in either case. Actually since you posted the original problem, it's safe to assume 'a' is positive.

Last edited: Apr 26, 2014
6. Apr 26, 2014

### Lengalicious

Yeah, that was my initial thought, just confuses me why the question uses z.. ooh well thanks I can actually get somewhere if I sub a in.

7. Apr 26, 2014

### Ray Vickson

You can look at the two series obtained by setting x = a and x = -a (both of which have |x|=a---assuming a > 0).

8. Apr 26, 2014

### Dick

The use of z instead of x is probably a typo. I'd ignore it. And your series is not complete. It should have an n=0 term. What is it?

9. Apr 26, 2014

Thanks guys!