Explicit Formula for Sum of Series

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
13 replies · 40K views
Andy111
Messages
26
Reaction score
0

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.
 
Physics news on Phys.org
Maybe I'm misunderstanding your question, but wouldn't it just be [tex]\sum_{n=1}^\infty 1/n[/tex] ?
 
The question asks for n terms, not an infinite number of terms
 
Oh, sorry. I read the original post too quickly.
 
That's alright, nothing to bash heads in about! As for the problem, it's a doozy...
 
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" , "there is no simple formula, akin to the formulae for the sums of arithmetic and geometric series, for the sum."
 
Last edited by a moderator:
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!
 
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.
 
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.
 
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.
 
That much is obvious. The trouble is finding a formula for the nth partial sum.
 
I think it's impossible, infinity would be the common denominator.
 
beanny007 said:
I think it's impossible, infinity would be the common denominator.
Not for the sum of n (a finite number) terms.
 
beanny007 said:
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.