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Power Series

  1. Oct 3, 2006 #1
    Suppose that [tex] \sum_{n=0}^{\infty} c_{n}x^{n} [/tex] converges when [tex] x=-4 [/tex] and diverges when [tex] x=6 [/tex]. What can be said about the convergence or divergence of the following series?

    (a) [tex] \sum_{n=0}^{\infty} c_{n} [/tex]

    (b) [tex] \sum_{n=0}^{\infty} c_{n}8^{n} [/tex]

    (c) [tex] \sum_{n=0}^{\infty} c_{n}(-3)^{n} [/tex]

    (d) [tex] \sum_{n=0}^{\infty} (-1)^{n}c_{n}9^{n} [/tex]

    So we know that [tex] \sum_{n=0}^{\infty} c_{n}x^{n} [/tex] converges when [tex] -5\leq x\leq5 [/tex], and diverges when [tex] x> 5 [/tex].

    (a) Would [tex] \sum_{n=0}^{\infty} c_{n} [/tex] diverge?
    (b) This would diverge because [tex] x>5 [/tex]?
    (c) This would converge, because [tex] -5<-3<5 [/tex]?
    (d) This would diverge because [tex] x>5 [/tex]?

  2. jcsd
  3. Oct 3, 2006 #2


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    This doesn't follow from what was given. To show the convergent series converge, all you need to know is that [itex]|c_n (-4)^n| \rightarrow 0 [/itex] as [itex]n \rightarrow \infty[/itex]. For the divergent ones, show that if a certain series [itex]a_n[/itex] diverges, then so does the series [itex]r^n a_n[/itex] whenever |r|>1.
  4. Oct 3, 2006 #3
    (a) [tex] \sum_{n=0}^{\infty} c_{n} [/tex] diverges because it [tex] \rightarrow \infty [/tex]?

    (b) [tex] \sum_{n=0}^{\infty} c_{n}8^{n} [/tex]. How would I use this [itex]|c_n (-4)^n| \rightarrow 0 [/itex] to establish that it converges?

    For the rest, they arent geometric series, right?
  5. Oct 3, 2006 #4


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    You have those backwards. Remember it's 4^n, not 1/4^n.
  6. Oct 5, 2006 #5
    So was I correct? Still not totally understanding it.

  7. Oct 5, 2006 #6


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    No. c_n 8^n grows faster than c_n 6^n, so the former diverges if the latter does.
  8. Oct 6, 2006 #7


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    If you are working with power series, you should have, long ago recognized that integers are not the only numbers that exist!

    Knowing that the series does not converge for x= 6 does NOT mean that it only converges for [itex]x\le 5[/itex]! It might converge for every x< 6.
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