Harmonic number divided by n

Question

How do I prove that $\dfrac{H_n}{n}$ (where $H_n$ is a harmonic number) converges to $0$, as $n \to \infty$?

Answer 1

Let $1\leqslant k\leqslant n$. Using the upper bounds $\frac1i\leqslant1$ for every $i\leqslant k$ and $\frac1i\leqslant\frac1k$ for every $k\lt i\leqslant n$ yields $H_n\leqslant k+\frac1k(n-k)=k-1+\frac{n}k$.

Choose $k$ such that $\sqrt{n}\leqslant k\leqslant \sqrt{n}+1$. Then $k-1\leqslant\sqrt{n}$ and $\frac1k\leqslant\frac1{\sqrt{n}}$ hence $H_n\leqslant2\sqrt{n}$. This is valid as soon as $\sqrt{n}\leqslant n$, that is, for every $n\geqslant1$.

Finally, for every $n\geqslant1$, $\frac{H_n}n\leqslant\frac2{\sqrt{n}}$, in particular $\frac{H_n}n\to0$ when $n\to\infty$.

Answer 2

$$ \frac{\dfrac12+\dfrac13+\dfrac14+\dfrac15+\cdots+\dfrac1n}{n} \le \frac{\displaystyle\int_1^n \frac{dx}{x}}{n} = \frac{\log n}{n} = \frac{m}{e^m}. $$ Every time $m$ increases by $1$, the numerator gets multiplied by a tiny amount when $m$ is big, and the denominator gets cut down to $1/e$ times what it was before, so the whole thing gets cut down to less than half what it was. So it approaches $0$.

(Then finally add the first term, $1$, in front of that series of fractions. That term is easy to deal with.)