Conc.interj."Grigore Moisil" Urziceni 2010 probl.1
Posted: Tue Mar 23, 2010 8:58 pm
Demonstrati ca pentru orice numar natural \( n\ge 2 \), are loc inegalitatea
\( \frac{1}{n+1}(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1})>\frac{1}{n}(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{2n}) \)
\( \frac{1}{n+1}(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1})>\frac{1}{n}(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{2n}) \)