Question

Cho \(\left(v_n\right)\left\{{}\begin{matrix}v_1=\dfrac{1}{2018}\\v_{n+1}=\dfrac{2v_n}{1+2018v_n^2},\forall n\in N^{\cdot}\end{matrix}\right.\)

CMR: \(v_{n+1}\ge v_n\)

Answer 1

Quy nạp 1 cách đơn giản, ta dễ dàng chứng minh dãy dương

Lại có: \(v_{n+1}=\dfrac{2v_n}{1+2018v_n^2}\le\dfrac{2v_n}{2\sqrt{1.2018v_n^2}}=\dfrac{1}{\sqrt{2018}}\)

\(\Rightarrow\) Dãy bị chặn trên bởi \(\dfrac{1}{\sqrt{2018}}\) hay \(v_n\le\dfrac{1}{\sqrt{2018}}\Leftrightarrow v_n^2\le\dfrac{1}{2018}\)  ; \(\forall n\ge1\)

\(\Leftrightarrow1-2018v_n^2\ge0\)

Ta có: \(v_{n+1}-v_n=\dfrac{2v_n}{1+2018v_n^2}-v_n=\dfrac{v_n-2018v_n^3}{1+2018v_n^2}=\dfrac{v_n\left(1-2018v_n^2\right)}{1+2018v_n^2}\ge0\)

\(\Rightarrow v_{n+1}\ge v_n\) (đpcm)