439. In the table of square roots, the difference between the roots of two consecutive numbers decreases, approaching zero, if the table is extended indefinitely. Why?
439. $\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$; the right side of this equality decreases, tending to zero, as it grows infinitely.

$$
\sqrt[3]{n+1}=\sqrt[3]{n}=\frac{1}{\sqrt[3]{(n+1)^{2}}+\sqrt[3]{(n+1) n}+\sqrt[3]{n^{2}}}
$$

from which

$$
\begin{gathered}
\frac{\sqrt[3]{n+1}-\sqrt[3]{n}}{\sqrt{n+1}-\sqrt{n}}= \\
=-\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt[3]{(n+1)^{2}}+\sqrt[3]{(n+1) n}+\sqrt[3]{n^{2}}}<\frac{2 \sqrt{n+1}}{3 \sqrt[3]{n^{2}}}<\frac{4 \sqrt{n}}{3 \sqrt[8]{n^{2}}}
\end{gathered}
$$

or

$$
\frac{4}{\sqrt[3]{n}}
$$

Therefore, the differences of the cube roots of natural numbers decrease more slowly than the differences of the square roots.