Harmonic series

Posted December 9th, 2007 by nico

We want to prove that harmonic series is divergent.
One elementary solution is an easy consequence of some well known inequality

$(1+\frac{1}{n})^{n}\le e\le(1+\frac{1}{n})^{n+1}$

This is equivalent to

$n\ln(1+\frac{1}{n})<1<(n+1)\ln(1+\frac{1}{n})$

$\frac{1}{n+1}<\ln\frac{n+1}{n}=\ln(n+1)-\ln n<\frac{1}{n}$

This inequality is also a consequence of Lagrange theorem applied to logarithmic function on interval [n,n+1].
Summing up from 1 to n we get

$\sum_{k=1}^{n}\frac{1}{k+1}<\sum_{k=1}^{n}\ln\frac{k+1}{k}=\sum_{k=1}^{n}\ln(k+1)-\ln k<\sum_{k=1}^{n}\frac{1}{k}$

$\sum_{k=1}^{n}\frac{1}{k+1}<\ln (n+1)<\sum_{k=1}^{n}\frac{1}{k}$

From the right side of the above relation we have

$\ln (n+1)<\sum_{k=1}^{n}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+...\frac{1}{n}$

Now let consider harmonic series

$\sum_{n=1}^{+\infty}\frac{1}{n}$

$\lim_{n \to +\infty}\ln (n+1)=+\infty$

taking limit in both sides we also have

$\lim_{n \to +\infty}1+\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+...\frac{1}{n}=+\infty$

As the partial sum of the harmonic series is a divergent sequence, the harmonic series is divergent.

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