olymmath

HMMT November 2008 T5

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An aside: the sum of all the unit fractions

It is possible to show that, given any real M , there exists a positive integer kk large enough that:

∑n=1k1n=11+12+13…>M\sum_{n=1}^{k} \frac{1}{n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3} \ldots>M

Note that this statement means that the infinite harmonic series, ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}, grows without bound, or diverges. For the specific example M=5\mathrm{M}=5, find a value of kk, not necessarily the smallest, such that the inequality holds. Justify your answer.

Dataset text NuminaMath olympiads. Quoted with source.

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