Problem statement here.
The problem was to show that fractional parts of harmonic numbers can get arbitrarily close to zero.
It is known that harmonic numbers are divergent. A quick proof of this can be done using the fact that $\frac{1}{x}$ is the derivate of $\log x$ and applying Rolle's theorem to get:
$$ \frac{1}{n+1} \lt \log(n+1) - \log n \lt \frac{1}{n}$$
Add up and get that harmonic number $H_n$ is at least as big as $\log (n+1) $ and so diverges.
Now let $n_k$ be the smallest integer such that $H_{n_k} \gt k$.
We must have that $H_{n_k - 1} \lt k \lt H_{n_k}$ i.e
$$ H_{n_k} - \frac{1}{n_k} \lt k \lt H_{n_k}$$
Thus the fractional part of $H_{n_k}$ is no more than $\frac{1}{n_k}$. Since $n_k \to \infty$ as $k \to \infty$, the fractional parts can be made arbitrarily small.
$n_k$ grows approximately as $e^{k - \gamma}$ where $\gamma$ is the euler mascheroni constant. It is known that $n_k$ is either $\lfloor e^{k - \gamma} \rfloor$ or $\lfloor e^{k - \gamma} \rfloor + 1$.
It is also conjectured to be the integer nearest to $e^{k - \gamma} + \frac{1}{2}$, attributed to Comtet (who proved the statement in the previous paragraph).
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