Inegalitate
Moderators: Bogdan Posa, Laurian Filip
Inegalitate
Sa se demonstreze ca \( \frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+...-\frac{1}{2005}+\frac{1}{2006}<0,2 \) .
. A snake that slithers on the ground can only dream of flying through the air.
-
Marius Mainea
- Gauss
- Posts: 1077
- Joined: Mon May 26, 2008 2:12 pm
- Location: Gaesti (Dambovita)
Fie \( S_1=\frac{1}{4}-\frac{1}{5}+...-\frac{1}{2005}=\frac{1}{4\cdot5}+\frac{1}{6\cdot7}+...+\frac{1}{2004\cdot2005} \) si
\( S_2=\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+
\frac{1}{2003\cdot2004} \)
Deoarece \( S_1<S_2 \)
\( 2S_1<S_1+S_2=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2004}-\frac{1}{2005}=\frac{1}{3}-\frac{1}{2005}=\frac{2002}{3\cdot2005} \)
Asadar \( S_1+\frac{1}{2006}<\frac{1001}{3\cdot2005}+\frac{1}{2006}<0,2 \)
\( S_2=\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+
\frac{1}{2003\cdot2004} \)
Deoarece \( S_1<S_2 \)
\( 2S_1<S_1+S_2=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2004}-\frac{1}{2005}=\frac{1}{3}-\frac{1}{2005}=\frac{2002}{3\cdot2005} \)
Asadar \( S_1+\frac{1}{2006}<\frac{1001}{3\cdot2005}+\frac{1}{2006}<0,2 \)