\(\frac{1}{u_{n+1}}=\frac{2\left(n+1\right)u_n+1}{u_n}=\frac{1}{u_n}+2n+2\)
\(\Leftrightarrow\frac{1}{u_{n+1}}-\left(n+1\right)^2-\left(n+1\right)=\frac{1}{u_n}-n^2-n\)
Đặt \(\frac{1}{u_n}-n^2-n=v_n\Rightarrow\left\{{}\begin{matrix}v_1=0\\v_{n+1}=v_n\end{matrix}\right.\)
\(\Rightarrow v_{n+1}=v_n=v_{n-1}=...=v_1=0\)
\(\Leftrightarrow\frac{1}{u_n}-n^2-n=0\Rightarrow u_n=\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
\(S_n=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}< \frac{2017}{2018}\)
\(\Leftrightarrow1-\frac{1}{n+1}< \frac{2017}{2018}\)
\(\Leftrightarrow...\)
