Bounds for the Harmonic k-th partial sum.

Solution 1:

For $m\ge 2$, $1/u\le 1/m \le 1/(u-1)$ for $u \in [m,m+1]$

By integration on this interval

$$\ln(m+1)-\ln m \le 1/m \le \ln m - \ln(m-1)$$

You then just have to sum those inequalities.

Solution 2:

Noting that $$ \frac1{x+1}<\frac1m<\frac{1}{x},x\in(m-1,m)$$ one has $$ \sum_{m=1}^k \frac1m=\sum_{m=1}^k\int_{m-1}^m\frac{1}{m}dx<1+\sum_{m=2}^k\int_{m-1}^m\frac{1}{x}dx=1+\int_1^k\frac1xdx=1+\ln k$$ and $$ \sum_{m=1}^k \frac1m=\sum_{m=1}^k\int_{m-1}^m\frac{1}{m}dx>\sum_{m=1}^k\int_{m-1}^m\frac{1}{x+1}dx=\int_1^k\frac1{x+1}dx=\ln (k+1). $$ So $$ \ln(k+1)<\sum_{m=1}^k \frac1m<1+\ln k. $$