Behaviour of series (radius of convergence)

In summary, the question asks for the behaviour of the series when z=-\rho_o. If z is non negative, the series is monatonic. However, if z is negative, the series is not monotonic.
  • #1
Lengalicious
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Homework Statement



Series:
[tex]\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{(x)^n}{na^n}[/tex]

what is the behaviour of the series at radius of convergence [tex]\rho_o=-z[/tex] ?


Homework Equations





The Attempt at a Solution


So I can specify that the series is monatonic if z is non negative as [tex]\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{(-z)^n}{na^n}[/tex] right?

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

Thanks, in advance!
 

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  • #2
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
Lengalicious said:

Homework Statement



Series:
[tex]\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{(x)^n}{na^n}[/tex]

what is the behaviour of the series at radius of convergence [tex]\rho_o=-z[/tex] ?


Homework Equations





The Attempt at a Solution


So I can specify that the series is monatonic if z is non negative as [tex]\sum_{n=1}^{\infty}(-1)^{(n+1)}\frac{(-z)^n}{na^n}[/tex] right?

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

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.
 
  • #4
Sorry for being unclear, I found the radius of convergence to be [tex]\rho_o=|a|[/tex] 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 [tex]z=-\rho_o[/tex]. I have added an image attachment of the question to clear things up hopefully.
 
Last edited:
  • #5
Lengalicious said:
Sorry for being unclear, I found the radius of convergence to be [tex]rho_o=|a|[/tex] 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 [tex]z=-\rho[\tex]. I have added an image attachment of the question to clear things up hopefully.

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.
 
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  • #6
Dick said:
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.
 
  • #7
Lengalicious said:
Sorry for being unclear, I found the radius of convergence to be [tex]\rho_o=|a|[/tex] 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 [tex]z=-\rho[\tex]. I have added an image attachment of the question to clear things up hopefully.

You can look at the two series obtained by setting x = a and x = -a (both of which have |x|=a---assuming a > 0).
 
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  • #8
Lengalicious said:
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?
 
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  • #9
Thanks guys!
 

Related to Behaviour of series (radius of convergence)

What is the radius of convergence of a series?

The radius of convergence of a series is a measure of how far from the origin the series can extend before it either converges or diverges. It is represented by the symbol "R" and is calculated using the ratio test or the root test.

How do you determine the radius of convergence of a series?

To determine the radius of convergence of a series, you can use the ratio test or the root test. The ratio test involves taking the limit of the absolute value of the ratio of the n+1th term to the nth term as n approaches infinity. If this limit is less than 1, the series converges. If it is greater than 1, the series diverges. The radius of convergence is then equal to the reciprocal of this limit. The root test involves taking the nth root of the absolute value of the nth term and then taking the limit as n approaches infinity. If this limit is less than 1, the series converges. If it is greater than 1, the series diverges. The radius of convergence is then equal to the reciprocal of this limit.

What happens if the radius of convergence is infinite?

If the radius of convergence is infinite, it means that the series converges for all values of x. This is also known as a power series with an infinite radius of convergence. In other words, the series can extend infinitely far from the origin in either direction and still converge.

What is the significance of the radius of convergence in power series?

The radius of convergence is important in power series because it tells us for which values of x the series will converge. If x is within the radius of convergence, the series will converge. If x is outside the radius of convergence, the series will diverge. This allows us to determine the interval of convergence for a power series and to approximate functions using power series.

Can the radius of convergence change?

Yes, the radius of convergence can change depending on the series. It is possible for a series to have a different radius of convergence for different values of x. For example, a series may have a radius of convergence of 3 for x<4 and a radius of convergence of 5 for x>4. Additionally, the radius of convergence can change if the series is manipulated in some way, such as by taking the derivative or multiplying by a constant.

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