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Find all real numbers such that the series converges

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Find all real numbers x ≠ −1 such that the series

    Ʃn=1 (1/n) . [(x-1)/(x+1)]n


    2. Relevant equations

    3. The attempt at a solution

    I really do not know where to start here. I assume you could first prove it converges sing a test but i'm really not sure.

  2. jcsd
  3. Nov 5, 2011 #2

    Well no, you're trying to find x such that the series converges, not prove that the series does converge, which it doesn't for many values of x.

    If I asked you to find the radius of convergence for [tex]\sum_{n=1}^{\infty}\frac{1}{n}y^{n}[/tex] would you know what I was on about? Or if you haven't done radius of convergence yet, would you be perhaps able to see which values of [itex]y[/itex] would make the above sum converge?
    Last edited: Nov 5, 2011
  4. Nov 5, 2011 #3
    Yes, i have brief experience with the radius of convergence, however I thought that more often than not you had to use the ratio test with the radius of convergence? In that example would the Radius = ∞?

    What values of y would we need for it to converge?

  5. Nov 5, 2011 #4
    Well if you have seen the radius of convergence then you should have seen the formula [tex]r=lim_{n\rightarrow\infty}|\frac{C_{n}}{C_{n+1}}|[/tex] so can you see how to use that to find the values of [itex]y[/itex]?

    And them presuming you can see it, do you see how to find the values of [itex]x[/itex] from that?
  6. Nov 5, 2011 #5

    Okay, using the formula you gave me:

    lim(n→∞) (n+1)/n . yn/yn+1

    = lim(n→∞) (n+1)/n . 1/y

    Now what?.. I can see the (n+1)/n diverges..
  7. Nov 5, 2011 #6
    No sorry I should have stated what the various things were.

    If you have something of the form [tex]\sum_{n=1}^{\infty}C_{n}y^{n}[/tex] then the radius of convergence, [itex]r[/itex] is defined by the formula I gave in the last post.

    And [itex]\frac{n+1}{n}[/itex] doesn't diverge.
  8. Nov 5, 2011 #7
    i don't understand. :/
  9. Nov 5, 2011 #8
    Okay so [tex]\sum_{n=1}^{\infty}\frac{1}{n}y^{n}[/tex] is of the form [itex]\sum_{n=1}^{\infty}C_{n}y^{n}[/itex] where [itex]C_{n}=\frac{1}{n}[/itex]

    Do you see now how to find the radius of convergence for the series?
  10. Nov 5, 2011 #9
    Okay.. so using that formula we get as n→∞ lim [ (n+1)/1 . 1/y ]

    but since (n+1)/1 converges to 1 then we have... 1/y < 1.. therefore y>1

    How's that
  11. Nov 5, 2011 #10
    No, you're not finding the limit of anything involving [itex]y[/itex], you're just finding [tex]r=\lim_{n\rightarrow\infty}\frac{C_{n}}{C_{n+1}}[/tex].

    And then from there you'll have [itex]-r<y<r[/itex] as values for which the series converges which you can use to find [itex]x[/itex] in your original question.
    Last edited: Nov 5, 2011
  12. Nov 5, 2011 #11
    Wow, i'm not functioning today..

    So Cn=1/n

    So, r = lim (1/n) / (1/n+1)

    r = lim (n+1)/n


    therefore, -1<y<1
  13. Nov 5, 2011 #12
    Yes! Now that you know the range of values y=(x-1)/(x+1) can take, you should be able to immediately see how to find the range of values x can take :)
  14. Nov 5, 2011 #13
    Thanks for your help.. I still cant see how I could use them.. Am i missing something here.. There are no values in my original question.
  15. Nov 5, 2011 #14
    Well you have that if you want [tex]\sum_{n=1}^{\infty}\frac{1}{n}y^{n}[/tex] to converge then you must have [itex]-4<y<4[/itex]

    Then you can simply say [itex]y=\frac{x-1}{x+1}[/itex] and so [itex]-4<\frac{x-1}{x+1}<4[/itex] and from there find the range of values [itex]x[/itex] can take for the original sum in your opening post to converge.
  16. Nov 6, 2011 #15
    Great. But why have you put the values for r = 4 ?? I thought we agreed that the r=1

    Therefore we would have -1<y<1 ?

    Thanks for everything
  17. Nov 6, 2011 #16
    Ah sorry I was very tired and mistyped, it should indeed have been -1<y<1.
  18. Nov 6, 2011 #17
    great no worries.

    Right so ive got it as x>0 and x>-1

    hows that?
  19. Nov 6, 2011 #18
    Almost right.
  20. Nov 6, 2011 #19
    Would it just be x>0.. since x≠-1..

    So x ε ℝ+ (ie.. positive real numbers)
  21. Nov 6, 2011 #20
    That's exactly right :)
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