Concursul "Congruente", problema 2

[Claudiu Mindrila]

a) Sa se arate ca \( \frac{1}{3}<\frac{1}{2k+1}+\frac{1}{2k+2}+...+\frac{1}{3k}<\frac{1}{2} \), oricare ar fi \( k\in \mathbb{N}, k\geq 2 \).
b) Fie \( n \in \mathbb{N}, n\geq 2 \). Demonstrati ca \( \frac{7}{12}< \frac{1}{2n+1}+\frac{1}{2n+2}+...+\frac{1}{4n}<\frac{5}{6} \).

Valentin Damian, Braila

[salazar]

a) \( \frac{1}{3k}<\frac{1}{2k+1} \)
\( \frac{1}{3k}<\frac{1}{2k+2} \)
..................
\( \frac{1}{3k}<\frac{1}{3k-1} \)
- nr. de term=\( 3k-2k-1+1=k \).
- deci \( S>\frac{1}{3k}\cdot k=\frac{1}{3} \)
\( S<\frac{1}{2k+1}\cdot k=\frac{k}{2k+1}<\frac{k}{2k}=\frac{1}{2}. \)

b) S=\( \frac{1}{2n+1}+...+\frac{1}{3n}+\frac{1}{3n+1}+...+\frac{1}{4n} \)
\( \frac{1}{3n}\cdot n=\frac{1}{3}<S_1<\frac{1}{2n+1}\cdot n<\frac{1}{2} \)
\( \frac{1}{4}<S_2<\frac{1}{3} \)
- insumand obtinem cerinta.