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

  • #1

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
 

Answers and Replies

  • #2
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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
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
Dick
Science Advisor
Homework Helper
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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
Dick
Science Advisor
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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
Dick
Science Advisor
Homework Helper
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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:

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