Solution:
We claim that:
$$A=\frac{1}{2015}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2015}\right)>B=\frac{1}{2016}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2016}\right) .$$
To prove this, let $S=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2015}$. Then $A=\frac{1}{2015} S$ and $B=\frac{1}{2016}\left(S+\frac{1}{2016}\right)$.
Thus, our proposed inequality can be written as:
$$\frac{1}{2015} S \stackrel{?}{>} \frac{1}{2016}\left(S+\frac{1}{2016}\right)$$
After multiplying both sides by $2015 \cdot 2016$ to clear some of the denominators, the proposed inequality becomes equivalent to:
$$2016 S \stackrel{?}{>} 2015 S+\frac{2015}{2016}$$
which, after subtracting $2015 S$ from both sides, is equivalent in turn to:
$$S \stackrel{?}{>} \frac{2015}{2016}$$
But $S>1$, so it follows that $S>\frac{2015}{2016}$, establishing the last inequality and thereby proving all of the previous inequalities. In particular, the proposed original inequality is correct:
$$A=\frac{1}{2015}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2015}\right)>B=\frac{1}{2016}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2016}\right) .$$