Cho dãy số \(\left(u_{n}\right)\), biết \(u_{i}=12, \frac{2 u_{n+1}}{n^{2}+5 n+6}=\frac{u_{n}+n^{2}-n-2}{n^{2}+n}\) với \(n \geq 1\). Tìm \(\lim \frac{u_{n}}{2 n^{2}+1}\).
Giải thích:
Ta có:
\(\begin{array}{l}\frac{2 u_{n+1}}{n^{2}+5 n+6}=\frac{u_{n}+n^{2}-n-2}{n^{2}+n} \Leftrightarrow \frac{2 u_{n+1}}{(n+2)(n+3)}=\frac{u_{n}}{n(n+1)}+\frac{n-2}{n} \\\Leftrightarrow \frac{2 u_{n+1}}{(n+1)(n+2)^{2}(n+3)}=\frac{u_{n}}{n(n+1)^{2}(n+2)}+\frac{n-2}{n(n+1)(n+2)}\end{array}\)\(\begin{array}{l}\Leftrightarrow \frac{2 u_{n+1}}{(n+1)(n+2)^{2}(n+3)}=\frac{u_{n}}{n(n+1)^{2}(n+2)}+\frac{2}{(n+1)(n+2)}-\frac{1}{n(n+1)} \\\Leftrightarrow \frac{u_{n+1}}{(n+1)(n+2)^{2}(n+3)}-\frac{1}{(n+1)(n+2)}=\frac{1}{2}\left[\frac{u_{n}}{n(n+1)^{2}(n+2)}-\frac{1}{n(n+1)}\right]\end{array}\)Đặt \(v_{n}=\frac{u_{n}}{n(n+1)^{2}(n+2)}-\frac{1}{n(n+1)}\), từ \((*)\) ta có \(v_{n+1}=\frac{1}{2} v_{n}\) nên \(\left(v_{n}\right)\) là cấp số nhân có công bội \(q=\frac{1}{2}, v_{1}=\frac{1}{2}\) suy ra \(v_{n}=v_{1} q^{n-1}=\frac{1}{2^{n}}\) \(\frac{u_{n}}{n(n+1)^{2}(n+2)}-\frac{1}{n(n+1)}=\frac{1}{2^{n}} \Leftrightarrow u_{n}=\frac{n(n+1)^{2}(n+2)}{2^{n}}+\left(n^{2}+3 n+2\right)\)
Khi đó
\(\lim \frac{u_{n}}{2 n^{2}+1}=\lim \frac{\frac{n(n+1)^{2}(n+2)}{2^{n}}+\left(n^{2}+3 n+2\right)}{2 n^{2}+1}=\lim \left[\frac{n(n+1)^{2}(n+2)}{2^{n}\left(2 n^{2}+1\right)}+\frac{n^{2}+3 n+2}{2 n^{2}+1}\right]\)Ta có \(2^{n}=C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+C_{n}^{3}+\ldots+C_{n}^{n}\gt C_{n}^{3}=\frac{n(n-1)(n-2)}{6}\)
Suy ra \(\lim \frac{n(n+1)^{2}(n+2)}{2^{n}\left(2 n^{2}+1\right)}=0\) và \(\lim \frac{n^{2}+3 n+2}{2 n^{2}+1}=\frac{1}{2}\)
Vậy \(\lim \frac{u_{n}}{2 n^{2}+1}=\frac{1}{2}\)
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