Behaviour of series (radius of convergence)

[Lengalicious]

Homework Statement

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 ?

Homework Equations

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 I am a bit confused because you cannot integrate (-1)^{(x+1)}\frac{(-z)^x}{xa^x}?

Thanks, in advance!

 

[vela]

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?

 

[Dick]

Homework Statement

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 ?

Homework Equations

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 I am a bit confused because you cannot integrate (-1)^{(x+1)}\frac{(-z)^x}{xa^x}?

Thanks, in advance!

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.

 

[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.

 

[Dick]

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[\tex]. I have added an image attachment of the question to clear things up hopefully.

<br /> <br /> 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 &#039;a&#039; to positive or negative. But the series will simplify a lot in either case. Actually since you posted the original problem, it&#039;s safe to assume &#039;a&#039; is positive.

 

[Lengalicious]

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.

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.

 

[Ray Vickson]

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[\tex]. I have added an image attachment of the question to clear things up hopefully.

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

 

[Dick]

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.

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?

 

[Lengalicious]

Thanks guys!