Any efficient way to calculate the sum of harmonic series upto nth term? 1 + 1/2 + 1/3 + --- + 1/n =?

Question

Is there any formula for this series "1 + 1/2 + 1/3 + --- + 1/n = ?" I think it is a harmonic number in a form of sum(1/k) for k = 1 to n.

@duffymo: Actually, math.stackexchange.com sounds like a perfect home for this question -- it's explicitly for "math at any level", unlike, say, mathoverflow.net. — Jim Lewis, Sep 20 '10 at 01:46
Thanks guys! I will visit that site. The answer of this question will help me solve a prob in Algorithm, which is computer-related though — user451587, Sep 20 '10 at 02:11
@Jim Lewis - Thanks for the heads up. I didn't realize that there were two math URLs now. — duffymo, Sep 20 '10 at 14:43
Integrate 1/x from 1 to n. Therefore it gives [ ln(x) + c ] as the answer. — Rahul, Dec 02 '12 at 18:42
Why not move it instead of closing it? Search engines still link to these closed questions and it's generally unproductive and unwelcoming to new users to treat them like this. — Xonatron, Jan 07 '14 at 19:04

You · Answer 1 · 2010-09-20T01:42:18.657

10

As it is the harmonic series summed up to n, you're looking for the nth harmonic number, approximately given by γ + ln[n], where γ is the Euler-Mascheroni constant.

For small n, just calculate the sum directly:

double H = 0;
for(double i = 1; i < (n+1); i++) H += 1/i;

edited Sep 20 '10 at 01:42

answered Sep 20 '10 at 01:35

You

score 3 · Answer 2 · answered Sep 20 '10 at 01:33

3

If I understood you question correctly, reading this should help you: http://en.wikipedia.org/wiki/Harmonic_number

answered Sep 20 '10 at 01:33

dee-see

score 2 · Answer 3 · answered Sep 20 '10 at 01:32

2

Here's one way to look at it:

answered Sep 20 '10 at 01:32

duffymo

score 0 · Answer 4 · answered Sep 20 '10 at 01:35

0

function do(int n) 
{
    if(n==1)
        return n;

    return 1/n + do(--n); 
}

answered Sep 20 '10 at 01:35

bevacqua

3

While recursive solutions look elegant, in this case it's inappropriate. – You Sep 20 '10 at 01:38
3

If the number is large enough you will get a stack overflow, or you will be adding basically zero, and not really changing the value much. – James Black Sep 20 '10 at 01:38
I figured he'd be using small sample numbers – bevacqua Sep 20 '10 at 01:40

4 Answers4