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Power Series

  • Thread starter mattmannmf
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  • #1
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Im given the following power series:
[tex]\sum[/tex] (x-3)^n/ n

I determined that the radius of convergence is R=1 and the interval of convergence is [2, 4)

They ask what values of x for which series converges absolutely?
and values of x for which series converges conditionally?

From what i read, my series should converge absolutely at |x-3|<1 (am i right?)

Im not sure about values of x where the series converges conditionally.
 

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  • #2
LCKurtz
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Im given the following power series:
[tex]\sum[/tex] (x-3)^n/ n

I determined that the radius of convergence is R=1 and the interval of convergence is [2, 4)

They ask what values of x for which series converges absolutely?
and values of x for which series converges conditionally?

From what i read, my series should converge absolutely at |x-3|<1 (am i right?)

Im not sure about values of x where the series converges conditionally.
Check the end points, x = 2 and 4.
 
  • #3
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yeah i did.
it converged on x=2
but diverged on x=4

so is that what they mean by absolute and conditionally?
 
  • #4
LCKurtz
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yeah i did.
it converged on x=2
but diverged on x=4

so is that what they mean by absolute and conditionally?
A series is conditionally convergent if is convergent but not absolutely convergent. So at x = 2, since it converges, you have to also check whether it converges absolutely. If it doesn't, then it is conditionally convergent.
 
  • #5
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I did the algebra and it came out that x=2 does converge abs.

so there are no other values that converge absolutely? what about in between the interval? do i have to check those too?

all other values outside the interval would diverge the series correct?
 
  • #6
LCKurtz
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I did the algebra and it came out that x=2 does converge abs.
We don't say that "x=2 does converge". We say "the series converges for x=2". But you better check again the "absolutely" part for x = 2.
so there are no other values that converge absolutely? what about in between the interval? do i have to check those too?

all other values outside the interval would diverge the series correct?
I thought you had already understood that you have absolute convergence inside the radius of convergence. And, of course, divergence beyond it.
 
  • #7
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well what i did to check the absolute convergence for x= 2 is I used the ratio test for:
E (x-3)^n/ n

And what I did was I just plugged in 2 for the x value and it came out to be -1 when I took the limit. And according to Ratio Test, <1 means abs conv.
 

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