 Solution

The function $1/x$ is decreasing and positive. Therefore on $[k,k+1]$ it is at most $1/k$, and the Riemann integral givesSumming from $k=1$ to $n-1$ yieldsFor $n=1$ both sides are zero; for $n>1$ the inequality is actually strict. The same integral test for convergence gives ${\sum }_{k=N}^{n-1}1/k\ge \log (n/N)$.