# Explicit Formula for Sum of Series

1. Apr 20, 2008

### Andy111

1. The problem statement, all variables and given/known data
Determine an explicit formula for the sum of n terms for the given series:

1, $$\frac{1}{2}$$, $$\frac{1}{3}$$, $$\frac{1}{4}$$, $$\frac{1}{5}$$

2. Relevant equations

3. The attempt at a solution

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

1, $$\frac{3}{2}$$, $$\frac{11}{6}$$, $$\frac{25}{12}$$, $$\frac{137}{60}$$

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

2. Apr 20, 2008

### steelphantom

Maybe I'm misunderstanding your question, but wouldn't it just be $$\sum_{n=1}^\infty 1/n$$ ?

3. Apr 20, 2008

### BrendanH

The question asks for n terms, not an infinite number of terms

4. Apr 20, 2008

### steelphantom

Oh, sorry. I read the original post too quickly.

5. Apr 20, 2008

### BrendanH

That's alright, nothing to bash heads in about! As for the problem, it's a doozy...

6. Apr 20, 2008

### steelphantom

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."

Last edited by a moderator: May 3, 2017
7. Apr 20, 2008

### sutupidmath

Well then wouldn't it just be:

$$\sum_{k=1}^n \frac{1}{k}$$

$$S_1=1$$

$$S_2=1+\frac{1}{2}=\frac{2+1}{2}$$

$$S_3=1+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}$$

$$S_4=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{12+6+4+3}{12}$$

blah, that doesn't actually seem to work!!

8. Apr 20, 2008

### BrendanH

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. Apr 20, 2008

### Dick

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. Mar 2, 2009

### blacklingo

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. Mar 2, 2009

### Tom Mattson

Staff Emeritus
That much is obvious. The trouble is finding a formula for the nth partial sum.

12. Feb 20, 2010

### beanny007

I think it's impossible, infinity would be the common denominator.

13. Feb 20, 2010

### Staff: Mentor

Not for the sum of n (a finite number) terms.

14. Feb 21, 2010

### icystrike

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

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