[m]\lim \limits_{k \to \infty} \frac{a_{k+1}}{a_k} = \lim \limits_{k \to \infty} (\frac{4^{k+5}}{5^{k+2}} : \frac{4^{k+4}}{5^{k+1}}) = \lim \limits_{k \to \infty} (\frac{4^{k+5}}{5^{k+2}} \cdot \frac{5^{k+1}}{4^{k+4}}) = [/m]
[m] = \lim \limits_{k \to \infty} (\frac{4^{k+5}}{4^{k+4}} \cdot \frac{5^{k+1}}{5^{k+2}}) = 4 \cdot \frac{1}{5} = \frac{4}{5}[/m]
Так как 4/5 < 1, то ряд сходится.