13. Which of these numbers is the smallest?
(A) $\frac{\sqrt{2018}}{2017}$
(B) $\frac{\sqrt{2016}}{2015}$
(C) $\frac{\sqrt{2019}}{2018}$
(D) $\frac{\sqrt{2017}}{2016}$
(E) $\frac{\sqrt{2020}}{2019}$
Problem 13. The answer is (E).

Let $n>1$. Since $n^{3}=n \cdot n^{2}>(n-1)\left(n^{2}-1\right)=(n-1)^{2}(n+1)$, it is easily seen that $n /(n-1)^{2}>(n+1) / n^{2}$ and thus that

$$
\frac{\sqrt{n}}{n-1}>\frac{\sqrt{n+1}}{n}
$$

Consequently, the smallest among the proposed numbers is $\sqrt{2020} / 2019$.

(Problem proposed by K. Kuzmin)