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Values for x which converge. (geometric series)

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the values of x for which the series converges. Find the sum of the series for those values of x.

    2. Relevant equations

    Sum (x+2)^n from n =1 to ∞

    3. The attempt at a solution

    OK so this is maybe fudging it.
    I said that the argument x+2 will diverge if the argument is >= 1. So I just messed with this.

    x+2 > = 1 ---> x >= -1 Values of x will diverge.

    Then to find the sum this is where it get shady I don't know what to do . Or if the above is OK.
  2. jcsd
  3. Nov 18, 2013 #2


    Staff: Mentor

    What about if x ≤ -3? Will your series converge or diverge?
  4. Nov 18, 2013 #3
    It will be ((-3)+2)^n = (-1)^n Diverge. Wouldn't it just oscillate between -1 and 1 depending on if n was even or odd?
  5. Nov 18, 2013 #4
    Wait I kind of read that wrong. I think it will converge when x < -3 except that you can't have x = -2?
  6. Nov 18, 2013 #5


    Staff: Mentor

    Yes, the series diverges. The terms of the series oscillate as you said. Using "it" to explain a concept that involves three things (series, sequence of partial sums, sequence of terms) does not lead to clarity.

    Write a few terms of your series when, say, x = -4. What does that series do?

    Why can't you have x = -2? What does the series look like in this case?
  7. Nov 18, 2013 #6
    Wait. Wait.

    |x+2| = x+2 >= 0 or -(x+2) x< 0 so

    x+2 < 1 for convergence so x < -1
    and -(x+2) < 1 implies x < - 3

    So it converges for -1 < x < -3 ?

    So for the sum then what do I do. If I just pretend that x is anything else and do the typical summation of geometric series a/ (1 - r) with a = (x+2) r = (x+2 )

    I get S = (x+2)/(-x-1) so do I just specify -1 < x < -3 ?
  8. Nov 18, 2013 #7


    Staff: Mentor

    A lot of what you wrote here makes no sense.
    No, that's not how the absolute value works.
    |x + 2| = x + 2 if x + 2 ≥ 0, and
    |x + 2| = -(x + 2) if x + 2 < 0.
    I think I know what you're trying to say, but you're not writing it correctly. -1 is not less than -3. It should be the series converges when -3 < x < -1.
    What do you mean "pretend that x is anything else"? The series converges for -3 < x < -1. If x is outside this interval, the series diverges, so why are you trying to find its sum?
    Last edited: Nov 18, 2013
  9. Nov 18, 2013 #8
    I thought this was what wiki said.
    If |x| = x if x>= 0 and -x if x<0 ?

    But what about the value for x = -2? Also I'm trying to find the sum because they books say " Find the sum of the series for those values of x.
    So that is what I was trying.
    Last edited by a moderator: Nov 18, 2013
  10. Nov 18, 2013 #9


    Staff: Mentor

    Right, but that's not what you wrote, which I've copied below.
    You must have been in a hurry, because you left out a lot in what you wrote.
    What does the series look like if x = -2? I asked this question several posts back (post #5), but you didn't respond.
  11. Nov 18, 2013 #10
    Sorry. I need to slow it down. I have 1000000000 things going on right now. OK so I'm not sure I understand what it would look like. It would be (0)^n = 0. I don't know what it looks like though .. Geometrically ?
  12. Nov 18, 2013 #11


    Staff: Mentor

    Write four or five terms of the series when x = -2. That's what I meant by "what does it look like?"
  13. Nov 18, 2013 #12
    Wouldn't it just be 0?

    (x+2)^n = (-2 +2) ^n + (-2 +2)^n + (-2+2)^n ?

    But what if X was just a tad bit less then -2 maybe -2.1 or -2.3. or in the other direction. I'm not sure what I'm supposed to think here. It gets smaller and smaller and smaller as it goes closer to two.

    Maybe its like -3 < x < -1 (x not equal to two).
    What am I supposed to do about the sum. Is the sum two then ? Because that is what it approaches?
  14. Nov 18, 2013 #13


    Staff: Mentor


    Your series is
    $$\sum_{n = 1}^{\infty}(x + 2)^n$$
    Write the first five terms of this series when x = -2.
    Please don't use the word "it", especially when "it" refers to two different things. Not only is that confusing to someone who is trying to understand what you're saying, but it also shows that things are very unclear in your mind. This is like saying, "He's coming over tonight, but I'll see him tomorrow" when the first "he" is Bob, but the "him" is Bill.
    Things aren't gelling here for you. You're still asking questions about things that you have already answered.

    What is the interval of convergence? You've already said that the series converges for -3 < x < -1, right? If x is any number in that interval, then the series converges. Period.

    The series is a geometric series, right? How do you find the sum of such a series for a particular value of x? You have a formula, right?

    What if the formula doesn't work for a particular x value, especially x = -2? You try something else, such as writing a few terms in the series. If you do this, you should immediately see what happens.
  15. Nov 18, 2013 #14
    ##\sum_{n = 1}^{\infty}((-2) + 2)^n = (0)^1 + (0)^2 + (0)^3 + ......##
  16. Nov 18, 2013 #15
    ##\sum_{n = 1}^{\infty}((-2) + 2)^n = (0)^1 + (0)^2 + (0)^3 + ......##

    I'm sorry I posted that before I was ready ignore my previous one.

    I don't understand. When x = -2 it is 0. I don't get what is the deal.

    I'm asking ...for the sum do I just use the formula and treat x as a number?

    ##\sum_{n = 1}^{\infty}(x + 2)^n = \sum_{n = 1}^{\infty}(x + 2)(x + 2)^{n-1} = (x-2)/( 1-(x +2)) = (x-2)/ (-x-1) ## Then do I just specify the interval ?

    Like -3 < x < -1 ...now this leads to my question about what to do when x = 2? Which brings us back to the top of this post.

    I'm sorry I'm trying here I don't understand this.
  17. Nov 18, 2013 #16

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    What are you not seeing? You have already told us that the interval of convergence is -3 < x < -1. The value x = 2 lies way outside the interval of convergence, so what does that tell you? The value x = -2 lies inside the interval of convergence, so what does that tell you?

    As for using the summation formula and then worrying about the interval: that is going about it backwards. FIRST decide on the interval of convergence, and THEN find a formula for the sum. For x outside the interval of convergence, the final sum formula does NOT represent the sum because the sum does not exist, but the formula does.
  18. Nov 18, 2013 #17
    It tells me x = 2 diverges because it is outside and it converges for x = -2. But when you evaluate it for x = -2 you just get 0 for the sum. Converges to 0?

    So is my formula correct and I just need to specify the values for which it works along side it?
  19. Nov 19, 2013 #18


    User Avatar
    Science Advisor
    Homework Helper

    The formula you quoted in post #15 is NOT correct. Why did you change the (x+2) you factored out into (x-2)????
  20. Nov 19, 2013 #19


    Staff: Mentor

    = 0, clearly. Was that so hard?
    x is a number, but the formula below is incorrect. It should give 0 when x = -2, but it doesn't.
  21. Nov 19, 2013 #20


    Staff: Mentor

    x = 2 neither converges nor diverges. x = 2 is just a number.
    What does? PLEASE DON'T SAY "IT"!
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