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Infinite Series convergence

  1. Apr 1, 2010 #1
    Okay, there's two questions, actually.

    First, determine if the series converges.

    SUM: (n-2)/(n^2-4n) (from n=5 to infinity)

    I used the integral test, found the integral to be 1/2 log(n^2-4n) from x=5 to x=t as t approaches infinity. That turned out to go to infinity so the series diverges.

    Does this seem right or did I do something wrong?

    Second, Suppose a series is the SUM: (1 - ((2^(k-1) - 1) / (2^(k+1)))

    Does the sum of the series converge or diverge? I'm not really even sure where to start this one.
  2. jcsd
  3. Apr 1, 2010 #2


    Staff: Mentor

    I used the ratio test and concluded the same thing.
    Here's what I think your series is:
    [tex]\sum_{k = 1}^{\infty} \frac{1 - 2^{k - 1} - 1}{2^{k + 1}}[/tex]

    If that's what it is, there's a lot of simplification you can do. After that you should be able to quickly determine whether it converges or not.
  4. Apr 1, 2010 #3
    No sorry, the first one is outside, then subtract the quantity starting with 2^k-1
  5. Apr 1, 2010 #4


    Staff: Mentor

    Like this?
    [tex]\sum_{k = 1}^{\infty} \left(1 - \frac{2^{k - 1} - 1}{2^{k + 1}}\right)[/tex]

    Try expanding the series to see what the first few terms look like.
  6. Apr 1, 2010 #5
    YEs, and I tried that. I did the limit as k went to infinity of 2^(k-1) / 2^(k+1) which I think is one, then the last bit is zero, so the total went to zero which doesn't help me much.
  7. Apr 1, 2010 #6
    That is incorrect. Try substituting a few values of k and you should immediately see what this term is. In general we have the rule:
    [tex]\frac{a^x}{a^y} = a^{x-y}[/tex]
    which you can use to calculate this term.

    Anyway if you want to show that a series [itex]\sum a_i[/itex] diverges a common way is to show that you can find a positive number M such that [itex]a_i \geq M[/itex]. In your specific example can you find a lower bound for
  8. Apr 1, 2010 #7
    Thank you, got it now.
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