Sum of an infinite series

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  • #1
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Homework Statement



∑ ( 60^(1/(n+3)) − 60^(1/(n+4)) )
n = 0

Homework Equations


I believe this is a geometric series so the sum would equal a/(1-r)


The Attempt at a Solution


I tried to view it as a geometric series but i had trouble finding a ratio, especially what i thought was r (60) would make the series diverge
 

Answers and Replies

  • #2
Dick
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You are right that it's not geometric. You aren't going to have much luck actually finding a sum. Do you just want to show it converges?
 
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  • #3
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Ever heard of the term telescopic sum? Try writing the sum up to n = 1, then n = 2, then for all n up to n = 3, and you will see what I mean.

(BTW: The answer is 60^(1/3) - 1.)
 
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  • #4
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yes but i'm not entirely sure how to do them. when I do that, I am left with 60^(1/3) and then 60^(1/infinity+3) - 60^(1/infinity+4) which would end up being 60^(1/3) since the others cancel but this is wrong. what am i doing wrong?
 
  • #5
Dick
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yes but i'm not entirely sure how to do them. when I do that, I am left with 60^(1/3) and then 60^(1/infinity+3) - 60^(1/infinity+4) which would end up being 60^(1/3) since the others cancel but this is wrong. what am i doing wrong?

Oh yeah, it does telescope in a manner of speaking. Thanks grey_earl. The terms in the split up series don't go to zero. You need to think of it as a limit of partial sums.
 
  • #6
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Exactly. Calculate first [tex]\sum_{n=0}^N (60^{1/(n+3)} -60^{1/(n+4)})[/tex], and then take the limit [tex]N \rightarrow \infty[/tex].
 
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