Question

Cmr ta luôn có : \(P_n=\frac{1.3.5.7.....\left(2n-1\right)}{2.4.6....2n}< \frac{1}{\sqrt{2n+1}}\) ( với mọi n ∈ \(Z^+\))

Answer 1

Ta có \(\left(2n\right)^2=4n^2>4n^2-1=\left(2n-1\right)\left(2n+1\right)\)

\(\Rightarrow\frac{1}{\left(2n\right)^2}< \frac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(P_n^2=\frac{1^23^25^2...\left(2n-1\right)^2}{2^24^26^2...2n^2}< \frac{1^23^25^2...\left(2n-1\right)^2}{1.3.3.5.5.7...\left(2n-1\right)\left(2n+1\right)}\)

\(P^2< \frac{1^23^25^2...\left(2n-1\right)^2}{1.3^2.5^2...\left(2n-1\right)^2\left(2n+1\right)}=\frac{1}{2n+1}\)

\(\Rightarrow P< \frac{1}{\sqrt{2n+1}}\)