Here is a completely elementary and self-contained proof (subject to accepting that $\frac{k}{2^k} \to 0$ as $k \to \infty$).
Let $a_n = \frac{1}{n}\sum_{k=1}^n \frac1{k} $. Then $na_n = \sum_{k=1}^n \frac1{k} $.
Therefore $\frac1{n+1} =(n+1)a_{n+1}-na_n =n(a_{n+1}-a_n)+a_{n+1} $ or $a_{n+1}-a_n =\frac1{n}(\frac1{n+1}-a_{n+1}) $.
Since, for $n > 1$, $a_n \ge \frac1{n}(1+\frac12) = \frac{3/2}{n} $, $a_{n+1}-a_n <-\frac{1}{2n(n+1)} $ so $a_n$ is a decreasing sequence.
We also have
$\begin{array}\\ (2n)a_{2n}-na_n &=\sum_{k=1}^{2n} \frac1{k}-\sum_{k=1}^n \frac1{k}\\ &=\sum_{k=n+1}^{2n} \frac1{k}\\ \text{so}\\ (2n)a_{2n}-na_n &< 1\\ \text{and}\\ (2n)a_{2n}-na_n &> \frac12\\ \end{array} $
Therefore $a_{2n} < \frac1{2n}(1+na_n) = \frac1{2n}+\frac12 a_n $ or
$\begin{array}\\ a_n &>2a_{2n}-\frac1{n}\\ &>2(2a_{4n}-\frac1{2n})-\frac1{n}\\ &=4a_{4n}-\frac1{n}-\frac1{n}\\ &=4a_{4n}-\frac{2}{n}\\ &=4(2a_{8n}-\frac1{4n})-\frac{2}{n}\\ &=8a_{8n}-\frac1{n}-\frac{2}{n}\\ &=8a_{8n}-\frac{3}{n}\\ \end{array} $
By induction we can show that $a_n > 2^ka_{2^kn}-\frac{k}{n} $ or $a_{2^kn} <2^{-k}(a_n+\frac{k}{n}) $.
Since both $\frac{k}{2^k} \to 0 $ and $\frac{a_n}{2^k} \to 0 $ as $k \to \infty$, $a_{2^kn} \to 0 $ as $k \to \infty$.
Since $a_n$ is decreasing and positive, $a_n \to 0$ as $n \to \infty$.