Summation of a Logarithmic Series

  • Thread starter S.R
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  • #1
S.R
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Homework Statement


What is the sum of the following series?

log(3/2)+log(4/3)+log(5/4)+...log(200/199).

Where log(x) is log base 10 of x.

Homework Equations





The Attempt at a Solution


Evidently, the previous form equals:

log(3/2*4/3*5/4*...200/199)

I'm missing something - no patterns are evident to me other than the denominator and numerator of the subsequent terms cancel out.

Any guidance would be appreciated.
 

Answers and Replies

  • #2
Curious3141
Homework Helper
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I'm missing something - no patterns are evident to me other than the denominator and numerator of the subsequent terms cancel out.

And that's important! What are you left with after cancelling *everything* that can be cancelled?
 
  • #3
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[tex]log(\frac{3}{2}*\frac{4}{3}*\frac{5}{4}*\frac{6}{5}..........*\frac{200}{199})[/tex]

Do you see in what pattern the terms cancel and which terms are left? :wink:
 
  • #4
S.R
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Intuitively, yes - would it be correct in saying that all terms cancel other than 1/2 and 200/1, leaving 200/2?
 
  • #5
Mentallic
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Intuitively no - however, would it be correct in saying that all terms cancel other than 1/2 and 200/1, leaving 200/2?
Yes, that would be correct, and you didn't find it intuitive even though you spotted the pattern?
 
  • #6
3,816
92
Intuitively no - however, would it be correct in saying that all terms cancel other than 1/2 and 200/1, leaving 200/2?

Yes, only 1/2 and 200/1 remains. 3 cancels 1/3, 4 cancels 1/4, 5 cancels 1/5 but there's no one to cancel 200 and 1/2.
 
  • #7
S.R
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Quickly edited after rereading :frown:. I'm not sure why I didn't recognize that pattern before.
 

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