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Sum of an infinite series(not quite geometric)

  • Thread starter GNelson
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Homework Statement



Determine the sum of the series:

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

Homework Equations





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]
 
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Answers and Replies

  • #2
rewriting it as a geometric series makes sense but how did 3k+2 become 32?
 
  • #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.
 
  • #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.
 
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  • #5
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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.
 
  • #6
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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?
 
  • #7
HallsofIvy
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Why would being "more of a mathematician" imply that you would be wrong about a math problem?:biggrin:
 
  • #8
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Why would being "more of a mathematician" imply that you would be wrong about a math problem?:biggrin:
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)
 
  • #9
dynamicsolo
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rewriting it as a geometric series makes sense but how did 3k+2 become 32?
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...
 
  • #10
Dick
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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...
No, it really is e^32. It's e^(3k+2) and the sum starts at k=10. That's e^(thirty two).
 
  • #11
dynamicsolo
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Oops, right: I misled myself in reading some of the other posts...
 

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