Explicit Formula for Sum of Series

  • Thread starter Andy111
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  • #1
Andy111
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Homework Statement


Determine an explicit formula for the sum of n terms for the given series:

1, [tex]\frac{1}{2}[/tex], [tex]\frac{1}{3}[/tex], [tex]\frac{1}{4}[/tex], [tex]\frac{1}{5}[/tex]


Homework Equations





The Attempt at a Solution



I calculated the first 5 terms for the sum sequence and got:

1, [tex]\frac{3}{2}[/tex], [tex]\frac{11}{6}[/tex], [tex]\frac{25}{12}[/tex], [tex]\frac{137}{60}[/tex]

but I can't find a pattern to determine an explicit formula.
 

Answers and Replies

  • #2
steelphantom
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Maybe I'm misunderstanding your question, but wouldn't it just be [tex]\sum_{n=1}^\infty 1/n[/tex] ?
 
  • #3
BrendanH
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The question asks for n terms, not an infinite number of terms
 
  • #4
steelphantom
159
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Oh, sorry. I read the original post too quickly.
 
  • #5
BrendanH
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That's alright, nothing to bash heads in about! As for the problem, it's a doozy...
 
  • #6
steelphantom
159
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You're right, it is a doozy! This series is known as a harmonic series, and according to http://plus.maths.org/issue12/features/harmonic/index.html" [Broken], "there is no simple formula, akin to the formulae for the sums of arithmetic and geometric series, for the sum."
 
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  • #7
sutupidmath
1,631
4
Well then wouldn't it just be:

[tex]\sum_{k=1}^n \frac{1}{k}[/tex]

[tex] S_1=1[/tex]

[tex]S_2=1+\frac{1}{2}=\frac{2+1}{2}[/tex]

[tex]S_3=1+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}[/tex]

[tex]S_4=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{12+6+4+3}{12}[/tex]

blah, that doesn't actually seem to work!
 
  • #8
BrendanH
63
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I know! It's whack!

I started out trying to find formulas for numerator and denominator separately. In cases where it doesn't simplify, the denominator is n! . As for the numerator, aside from the obvious (n! + n!/2 + n!/3+...+n!/n) I do not see how to arrive at a formula.
 
  • #9
Dick
Science Advisor
Homework Helper
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According to http://mathworld.wolfram.com/HarmonicSeries.html you can write it as a sum of the Euler-Mascheroni constant and a digamma function. I'm guessing if there were an easier expression they would have mentioned it. I think steelphantom is right, there is no elementary formula.
 
  • #10
blacklingo
1
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I think all it is is sigma from k=2 to n of 1/n. very simple that would start off with one half then one third then one quarter and so on.
 
  • #11
quantumdude
Staff Emeritus
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5,575
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That much is obvious. The trouble is finding a formula for the nth partial sum.
 
  • #12
beanny007
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I think it's impossible, infinity would be the common denominator.
 
  • #13
36,477
8,439
I think it's impossible, infinity would be the common denominator.
Not for the sum of n (a finite number) terms.
 
  • #14
icystrike
446
1
I think it's impossible, infinity would be the common denominator.

The denominator ought to be the lcm(1,2,...,n) whereas the numerator is the puzzle.
 

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