\(u_n:\left\{{}\begin{matrix}u_1=1\\u_{n+1}=3u_n+2n-1\left(1\right)\end{matrix}\right.\)
Đặt \(limu_n=a\Rightarrow limu_{n+1}=a\)
\(\left(1\right)\Rightarrow a=3a+2n-1\)
\(\Rightarrow a=\dfrac{1-2n}{2}\)
\(\Rightarrow limu_n=\dfrac{1-2n}{2}\)
\(\Rightarrow lim\dfrac{u_n}{3^n}=lim\dfrac{1-2n}{2.3^n}=0\)
Đặt \(v_n=u_n+n\)
Chứng minh được \(3^n>n^2\) với mọi số nguyên dương n bằng phương pháp quy nạp. Suy ra: \(\left|\dfrac{n}{3^n}\right|< \left|\dfrac{n}{n^2}\right|=\dfrac{1}{n}\). Mà \(lim\dfrac{1}{n}=0\rightarrow lim\dfrac{n}{3^n}=0\)
\(u_{n+1}=3u_n+2n-1\rightarrow v_{n+1}=3v_n\\ \rightarrow v_n=v_1.3^{n-1}=2.3^{n-1}\\ \rightarrow u_n=2.3^{n-1}-n\\ lim\dfrac{u_n}{3^n}=lim\dfrac{2.3^{n-1}-n}{3^n}=lim\left(\dfrac{2}{3}-\dfrac{n}{3^n}\right)=\dfrac{2}{3}\)
