What is the sum of reciprocals of Natural Numbers?

Asked Apr 05 '15 at 12:14

Active Apr 05 '15 at 12:14

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I want to calculate the sum of first $n$ natural numbers. I used the following C program to compute the first '$n$' digits :

#include <stdio.h>
int main() {
    int n;
    double m;
    double sum=0;
    scanf("%d", &n);
    while (n>0) {
        m=1.0/n;
        sum+=m;
        n--;
    }
    printf("%lf",sum);
    return(0);
}

What I get is that the sum is very slowly diverging.

For $10^{1}$, I get the value : $2.92...$
For $10^{14}$, I get the value : $18.807..$
For $10^{16}$, I get the value : $21.92..$

So it appears to diverge.

How can we prove it? Actually sum of reciprocals is divergent, even after being a subseries of this series, so it naturally should be divergent. How can we show this?

asked Apr 05 '15 at 12:14

Da real Ninja

So, you want to show that $\displaystyle\sum_{i=1}^\infty \dfrac{1}{i}$ diverges? Right? – Prasun Biswas Apr 05 '15 at 12:16
http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#Divergence – lab bhattacharjee Apr 05 '15 at 12:16
Oh thank you lab battacharjee! lol my computer is still computing 10^24 ! – Da real Ninja Apr 05 '15 at 12:18
@labbhattacharjee Thats a very simple proof! – Da real Ninja Apr 05 '15 at 12:19

What is the sum of reciprocals of Natural Numbers?

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