Solution
= Solution
The <function> $1/x$ is decreasing and positive. Therefore on $[k,k+1]$ it is at most $1/k$, and the <Riemann integral> gives
$$
\int_k^{k+1}\frac{dx}{x}\le\frac1k.
$$
Summing from $k=1$ to $n-1$ yields
$$
\boxed{\log n=\int_1^n\frac{dx}{x}\le\sum_{k=1}^{n-1}\frac1k.}
$$
For $n=1$ both sides are zero; for $n>1$ the inequality is actually strict. The same <integral test for convergence> gives $\sum_{k=N}^{n-1}1/k\ge\log(n/N)$.