[tex] Din\ \frac{k}{(k+1)^2}< \frac{1}{k} \ deducem: \\
\frac{1}{2^2}+ \frac{2}{3^2}+ \frac{3}{4^2}+...+ \frac{63}{64^2} < \frac{1}{1} +2\cdot \frac{1}{2} +4\cdot \frac{1}{4} +8\cdot \frac{1}{8} +16\cdot \frac{1}{16}+32\cdot \frac{1}{32} =6\\
deoarece: \frac{1}{1} = \frac{1}{1} ,\frac{1}{2} = \frac{1}{2},\frac{1}{3} <\frac{1}{2},\frac{1}{4} = \frac{1}{4},\frac{1}{5} <\frac{1}{4},\frac{1}{6} <\frac{1}{4},\frac{1}{7} <\frac{1}{4},\frac{1}{8} = \frac{1}{8},...[/tex]
