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Radius of convergence of an infinite summation

  • Thread starter jacobrhcp
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  • #1
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[SOLVED] radius of convergence of an infinite summation

Homework Statement



find the radius of convergence of the series:

[tex]\sum\frac{(-1)^k}{k^2 3^k}(z-\frac{1}{2})^{2k}[/tex]

Homework Equations



the radius of convergence of a power series is given by [tex]\rho=\frac{1}{limsup |c_k|^{1/k}}[/tex]

and is equal to [tex]\frac{1}{R}[/tex] when [tex]Lim_{n->\infty} \frac{|c_{k+1}|}{|c_k|}= R [/tex]

The Attempt at a Solution



the major thing I'm stuck on is the '2k' in the series. General power series only have a factor 'k' in their powers, and I don't know how to get rid of it, so I can use the formula for radius of convergence of power series.
 

Answers and Replies

  • #2
Defennder
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Well, actually if you did the ratio test you'll notice that 2k (both numerator and denominator_ will cancel itself out.
 
  • #3
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you're quite right.

I am not used to the tests of general infinite series, because we did power series before we handeled general infinite series and it's tests (such as the ratio test), which I'm afraid we won't ever cover. =( I'll have to study them by myself.

Thanks.
 
  • #4
arildno
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Remember the following, jacob:

We have basically, only one tool to determine whether a series converges or not:

Its "relation" to a geometric series!

Practically all of the so-called tests work because in the proof of them, we are able to tweak the condition into a comparison with a relevant geometric series.
 
  • #5
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how is that true? I mean, there are other infinite summations than the geometric series for which we know the radius of convergence, right?
 
  • #6
arildno
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Sure, there are!
But just about every test we have shows, in the proof of it, that if the test criterion holds, then our actual series will have some unspecified limit below that of a convergent geometric series.
 

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