Finding the interval of convergence and the radius of a power series

In summary: But for the given series, the limit comparison test is easier.In summary, the power series given converges for all values of x where |x| < 1, and the radius of convergence is 1. The series does not converge at x = 1 or x = -1, but it does converge at any value in between. The convergence can also be proven using the integral test or the limit comparison test.
  • #1
Ratio Test =)
19
0

Homework Statement


Find the radius and interval of convergence of the power series:
[tex]\sum_{n=1}^{\infty} \frac{x^{2n-1}}{(n+1)\sqrt{n}}[/tex]


Homework Equations


..


The Attempt at a Solution


My soltion:
the ratio test will gives |x^2|=|x|^2
it converges if |x|^2 < 1
i.e. if |x|<1
i.e. if -1<x<1
But the problem here when i substitute x=1 and x=-1 it will give the same series
Is this right?
The resulting series converges by Integral Test.
Radius = 1
 
Physics news on Phys.org
  • #2
x = - 1 doesn't give the same series as x = 1. If x = -1, the series is
[tex]\sum_{n=1}^{\infty} \frac{(-1)^{2n-1}}{(n+1)\sqrt{n}}[/tex]
Since 2n - 1 is odd for all integers n, the numerator of the expression in the summation is always -1. If x = 1, the numerator is always +1.
 
  • #3
Mark44 said:
x = - 1 doesn't give the same series as x = 1. If x = -1, the series is
[tex]\sum_{n=1}^{\infty} \frac{(-1)^{2n-1}}{(n+1)\sqrt{n}}[/tex]
Since 2n - 1 is odd for all integers n, the numerator of the expression in the summation is always -1. If x = 1, the numerator is always +1.

Oh...Thanks
But will not affect the convergence right?
since the series which obtained by multiplying a convergent series by a constant is still convergent.
Right?
 
  • #5
No No Sorry!
there was a typo!
its 2n-2 not 2n-1 in the series =)
Sorry Again.
 
  • #6
Ratio Test =) said:
No No Sorry!
there was a typo!
its 2n-2 not 2n-1 in the series =)
Sorry Again.

Ok, then it is the same series at x=1 and x=(-1). So you only have to check one of them. Is it convergent?
 
  • #7
Thanks you.
you saved my life !
It is easy for me.
It converges by integral test.
 
  • #8
Ratio Test =) said:
Thanks you.
you saved my life !
It is easy for me.
It converges by integral test.

Sure. Or you could compare it to the convergent p-series 1/n^(3/2).
 
  • #9
Ohhh
My bad
How in Earth I did not notice this :@
 
  • #10
Ratio Test =) said:
Ohhh
My bad
How in Earth I did not notice this :@

I didn't say your way was bad. The integral test works fine.
 
Last edited:

FAQ: Finding the interval of convergence and the radius of a power series

What is the interval of convergence for a power series?

The interval of convergence for a power series is the range of values for which the series converges. It can be found by using various convergence tests such as the ratio test or the root test.

How do you calculate the radius of convergence?

The radius of convergence is calculated by finding the limit of the ratio of consecutive terms in the power series. This limit is known as the convergence ratio or the radius of convergence. The series will converge within this radius and diverge outside of it.

Can a power series have an infinite radius of convergence?

Yes, it is possible for a power series to have an infinite radius of convergence. This means that the series will converge for all values of the variable. This is usually the case when the series is a polynomial or a geometric series.

How do you determine the endpoints of the interval of convergence?

The endpoints of the interval of convergence can be determined by plugging in the endpoints of the interval into the original power series. If the series converges for both endpoints, then they are included in the interval of convergence. If the series diverges for one or both endpoints, then they are not included in the interval of convergence.

Are there any other methods for finding the interval of convergence and the radius of a power series?

Yes, there are other methods such as the integral test and the comparison test that can also be used to determine the interval of convergence and the radius of a power series. However, these tests are not as commonly used as the ratio test or the root test.

Similar threads

Interval of convergence of power series from ratio test
Replies
2
Views
819
On the ratio test for power series
Replies
2
Views
1K
Pointwise convergence for all real x
Replies
5
Views
366
How to show that these sequences are summable?
Replies
1
Views
1K
Proving convergence and divergence of series
Replies
3
Views
846
Intervals of Convergence- Power Series
Replies
11
Views
2K
Radius of Convergence and Power Series Expansion Around Multiple Values
Replies
6
Views
1K
On the definition of radius of convergence; a small supremum technicality
Replies
7
Views
1K
Set of convergence of a Power series
Replies
4
Views
2K
Addition of power series and radius of convergence
Replies
14
Views
2K

Hot Threads

Recent Insights

Back
Top