Proving the harmonic series, multiplied by a factor of 1/n, decreases monotically to zero. I would like to show that $\left(1 + \frac 12 + \dots + \frac 1n\right) \cdot \left(\frac 1n\right)$ decreases monotonically to zero.
I have seen one method: to first show that the difference $\left(1 + \frac 12 + \dots + \frac 1n\right) - \log(n)$ decreases monotonically to the Euler-Mascheroni constant, but I was wondering if there is another (perhaps cleaner or more instructive) way of proving the limit is zero.
Any help would be greatly appreciated.
Thanks,

Hint
For large values of $n$, the expansion of $H_n$ is $$H_n=\gamma +\log (n)+\frac{1}{2 n}-\frac{1}{12
   n^2}+\frac{1}{120 n^4}+O\left(\left(\frac{1}{n}\right)^6\right)$$ which leads to $$\frac{H_n}{n}=\frac{\gamma +\log (n)}{n}+\frac{1}{2 n^2}-\frac{1}{12
   n^3}+\frac{1}{120 n^5}+O\left(\left(\frac{1}{n}\right)^7\right)$$
