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Testing convergence of Series

  1. May 24, 2015 #1
    1. The problem statement, all variables and given/known data

    "Determine whether the following series converge or diverge. If the series is geometric or telescoping, find its sum.":

    ## \left ( \sum_{k=1}^\infty2^{3k} *3^{1-2k} \right)##

    2. Relevant equations

    The different tests for convergence?

    3. The attempt at a solution

    Ok, I've looked at all the tests for convergence I know, and as far as I've been able to tell, none of them can work with this? Here's a quick outline about each of the tests and why I don't think I can use it.

    Divergence Test: I have to say I can't even do the limit of this. It just gives me ##2^{infinity}*3^{1-infinity}##, no?
    Geometric series: I don't have only one exponent, so I can't do this. Neither of the exponents looks like ##(n-1)## so I don't know.
    Telescoping series: There's nothing that cancels out. I also can't take the limit.
    p-Series: It's not in the form ##{\frac {1} {p}}##.
    Alternating series: It doesn't have a ##-1^n## term.
    Integral test: ##\int 2^{3k} *3^{1-2k} \, dx## . I can't integrate this because of the ##k## being a power.
    Root test: While both are raised to a power of ##k##, I don't think I can apply the root test because of the ##1-2k##.
    Ratio test: This is usually my go to, and I thought it would work. However, I get ##\frac {1} {9}## and then the ##3^{1-2k}## just stays there.
    Comparison test: Even if I wanted to use the limit or direct comparison, I wouldn't even know what to compare it with.

    The solutions tell me it's supposed to converge to 24 and to use the geometric test. However, I really don't see how the geometric test is supposed to apply.

    Thanks for your time, and I hope I did everything right in posting this!
     
  2. jcsd
  3. May 24, 2015 #2

    wabbit

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    Gold Member

    This is simpler than what you're trying. Go back to the expression under the sum and try to see if you can put it in a simpler form first. Use the relations you know about exponents and multiplication, such as ## x^{nk}=(x^n)^k ##, or ## x^{p+q}=x^px^q ##, etc.
     
  4. May 24, 2015 #3
    Ooh, that helped!
    Ok, so doing that I get ##(2^3)^k*3^1*3^{-2k}##. That becomes ##3*8^k*3^{-2k}##. I can see where the 24 is supposed to come from: 3*8. I just can't get it to simplify to that.
    I rearrange the expression so it's ##\frac{3*8^k}{3^{2k}}## and then that becomes ##({\frac{8}{9}})^k *3## If the summation started at k=0, I would see that this is a geometric sum with a = 3, r = 8/9. However, the summation starts at k=1, so shouldn't I have ##n-1## as an exponent?
     
    Last edited: May 24, 2015
  5. May 24, 2015 #4

    wabbit

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    Gold Member

    Yep, you got it - about the starting point ## k=1 ## , you just need to be careful when determining the actual sum, using things like ## \sum_{k\geq 1}a_k=(\sum_{k\geq 0}a_k)-a_0 ##, or in your case using ## a^k=a\cdot a^{k-1} ## : while your series may not be exactly the standard geometric sum, you can express in terms of that.
     
  6. May 24, 2015 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Either change the exponent or subtract off the "k= 0" term.
     
  7. May 24, 2015 #6
    Alright, thank you very much to the two of you, this really helped me out!

    Have a great day! :smile:
     
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