Finding the radius of convergence of a power series

In summary, the problem is that the power series needs to be in the form x^(3n+1), not x^(3n). This can be fixed by setting t=x^(3n+1).
  • #1
R Letter
2
1

Homework Statement


Σ(n=0 to ∞) ((20)(-1)^n(x^(3n))/8^(n+1)

Homework Equations


Ratio test for Power Series: ρ=lim(n->∞) a_(n+1)/a_n

The Attempt at a Solution


I tried the ratio test for Power Series and it went like this:

ρ=lim(n->∞) (|x|^(3n+1)*8^(n+1))/(|x|^(3n)*8^(n+2))
=20|x|/8 lim(n->∞) 1
=20|x|/8

20|x|/8<1
|x|<2/5

So radius of convergence=2/5. However when I input the power series into Mathematica, it says that the radius of convergence should be 2 (http://goo.gl/9nAHoS)

Where am I going wrong in my calculation?
 
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  • #2
Hi there mate, I am going to attempt to help.

Here are some possible errors that I am looking into:

First, x^(3n) when you make it n+1 in the numerator of the ratio test, it should be x^(3n+3), correct?

Well, that might do it. Does this help?
 
  • #3
R Letter said:

Homework Statement


Σ(n=0 to ∞) ((20)(-1)^n(x^(3n))/8^(n+1)

Homework Equations


Ratio test for Power Series: ρ=lim(n->∞) a_(n+1)/a_n

The Attempt at a Solution


I tried the ratio test for Power Series and it went like this:

ρ=lim(n->∞) (|x|^(3n+1)*8^(n+1))/(|x|^(3n)*8^(n+2))
=20|x|/8 lim(n->∞) 1
=20|x|/8

20|x|/8<1
|x|<2/5

So radius of convergence=2/5.However when I input the power series into Mathematica, it says that the radius of convergence should be 2 (http://goo.gl/9nAHoS)

Where am I going wrong in my calculation?
Try setting ##x^3 = t## and finding the radius of ##t##-convergence for ##\sum_n c_n t^n##. Then translate those results into statements about ##x##. Alternatively, do it over again, but repair the algebraic errors you made. (Avoidance of such errors is the reason I suggested looking at ##t## instead of ##x##.)
 
  • #4
Also, one other thing --> In your solution you have a 20 sticking out, however, it seems to me that when you take the a^(n+1) term and the a^n term and divide them, the 20 on each will cancel leaving |x^3|/8 <1
 
  • #5
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Likes RJLiberator

What is the definition of the radius of convergence of a power series?

The radius of convergence of a power series is a value r that represents the distance from the center of the power series in which the series converges. In other words, if the distance from the center to a specific point is less than r, the series will converge at that point.

How do you find the radius of convergence of a power series?

To find the radius of convergence, you can use the Ratio Test, which compares the absolute value of a term in the series to the absolute value of the next term. If the limit of this ratio is less than 1, the series converges. The radius of convergence is then given by the reciprocal of this limit.

What is the significance of the radius of convergence?

The radius of convergence is important because it tells us the interval of values for which the power series will converge. If a point is within this interval, the series will converge at that point. If a point is outside of this interval, the series will diverge at that point.

Can the radius of convergence be negative?

No, the radius of convergence cannot be negative. It is always a positive value that represents the distance from the center of the power series.

How does the radius of convergence relate to the convergence of a power series?

The radius of convergence is directly related to the convergence of a power series. If the distance from the center to a point is less than the radius of convergence, the series will converge at that point. If the distance is greater than the radius of convergence, the series will diverge at that point.

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