A Clever Trick to Prove the Divergence of the Harmonic Series

The inequality \(e^x \geq 1+x\) offers a neat way to show that the harmonic series diverges. Applying it for \(x = \frac{1}{n}\) we get:

\[e^{1/n} \geq 1 + 1/n = \frac{n+1}{n}.\]

Repeating this idea \(n\) times gives:

\[e^{1 + 1/2 + 1/3 + \dots + 1/n} \geq \frac{2}{1} \cdot \frac{3}{2} \cdot \dots \cdot \frac{n+1}{n} = n+1.\]

Taking \(log\) both sides shows that the harmonic sum \(H_n := 1 + 1/2 + \dots + 1/n\) satisfies:

\[H_n \geq \log(n+1),\]

proving that the harmonic series diverges.

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Inspired by Ragib Zaman’s comment on Math Stack Exchange, June 12, 2014.