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Sum of an infinite series(not quite geometric)

  1. Feb 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine the sum of the series:

    [tex]\sum[/tex][tex]^{infinity}_{K=10}[/tex] [tex]\frac{7}{e^(3k+2)}[/tex]

    2. Relevant equations



    3. The attempt at a solution


    limit n->infinity of sn=[tex]\sum[/tex][tex]^{n}_{K=10}[/tex] [tex]\frac{7}{e^(3k+2)}[/tex]=[tex]\frac{7}{e^(32)}[/tex]+[tex]\frac{7}{e^(35)}[/tex]....[tex]\frac{7}{e^(3n+2)}[/tex]

    This series does not exactly fit a geometric series or any partial fraction I can reduce. After this point I am stuck, any hints are greatly welcome.

    My alternative method I attempted was to try to re-write it in such a way it was geometric., where f=1 is when k=10
    that is that [tex]\sum[/tex][tex]^{infinity}_{K=10}[/tex] [tex]\frac{7}{e^(3k+2)}[/tex]= [tex]\sum[/tex][tex]^{infinity}_{f=1}[/tex][tex]\frac{7}{e^(32)}[/tex]*([tex]\frac{1}{e^(3)}[/tex])^(f-1) which when simplfied I got = [tex]\frac{7}{e^(31)-1}[/tex]
     
    Last edited: Feb 10, 2008
  2. jcsd
  3. Feb 10, 2008 #2
    rewriting it as a geometric series makes sense but how did 3k+2 become 32?
     
  4. Feb 10, 2008 #3

    quasar987

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    Have you investigated the consequences of [tex]e^{3k+2} = e^{2}(e^3)^k[/tex]?

    P.S. You want to write [ t e x ] \sum_{k=10}^{\infty} [ \ t e x ]

    and

    [ t e x ] e^{3k+2} [ t e x ]

    for it to come out correctly.
     
  5. Feb 10, 2008 #4

    HallsofIvy

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    quasar987's point is that this is not "not quite" but exactly a geometric series. There is a standard formula for the sum of a geometric series.
     
    Last edited: Feb 10, 2008
  6. Feb 10, 2008 #5
    My rational in the re-write was to write it in such a way that the terms of the sequence would be identical, in the orignal form we obtain 7/e^32+7/e^35 ect.. which was the only method of solution that I could see that had a chance of working. I tested each k value against f values at f=1 against k=10, they produce identical terms for every valid k the only difference being the new series starts at 1 rather than 10 but the way I see it we are summing an infinite amount of terms,I was curious if this was mathematically valid.
     
  7. Aug 7, 2008 #6
    Correct me if I'm wrong (I'm more of a mathematician really), and I'm not all that familiar with your notation, but it is a convergent geometric progression.

    It can be rearranged as sum to infinity of 7e^-2 * (e^-3)^k

    which is a geometric progression with the first term (a) of 7e^-32

    and increases by a factor (r) of e^-3

    Using the geometric progression formula of summation of series i.e. sum=a/(1-r)

    sum= (7e^-32)/(1-e^-3)

    Isn't it?
     
  8. Aug 7, 2008 #7

    HallsofIvy

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    Why would being "more of a mathematician" imply that you would be wrong about a math problem?:biggrin:
     
  9. Aug 7, 2008 #8
    In itself, it wouldn't. But, as I mentioned before I haven't used notation quite like that, so my interpretation could be wrong. (I assumed that the notation was specific to physics)
     
  10. Aug 8, 2008 #9

    dynamicsolo

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    That's intended to be the staged exponential [tex](e^3)^{2}[/tex] . One of the things TeX doesn't seem to do is double exponents. The intent of the expression is clear, if the notation may not be...
     
  11. Aug 8, 2008 #10

    Dick

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    No, it really is e^32. It's e^(3k+2) and the sum starts at k=10. That's e^(thirty two).
     
  12. Aug 9, 2008 #11

    dynamicsolo

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    Oops, right: I misled myself in reading some of the other posts...
     
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