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\(u_{n+1}=\dfrac{u_n}{u_n+1}\Rightarrow\dfrac{1}{u_{n+1}}=\dfrac{1}{u_n}+1\)
Đặt \(\dfrac{1}{u_n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{u_1}=1\\v_{n+1}=v_n+1\end{matrix}\right.\)
\(\Rightarrow v_n\) là CSC với công sai \(d=1\Rightarrow v_n=v_1+\left(n-1\right).1=n\)
\(\Rightarrow u_n=\dfrac{1}{n}\)
\(\Rightarrow u_n+1=\dfrac{n+1}{n}\)
\(\lim\dfrac{2014\left(\dfrac{2}{1}\right)\left(\dfrac{3}{2}\right)\left(\dfrac{4}{3}\right)...\left(\dfrac{n+1}{n}\right)}{2015n}=\lim\dfrac{2014\left(n+1\right)}{2015n}=\dfrac{2014}{2015}\)
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\(\left\{{}\begin{matrix}u_1=a;u_2=b\\u_{n+2}=\dfrac{1}{2}u_{n+1}+\dfrac{1}{2}u_n\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u_1=a,u_2=b\\u_{n+2}+\dfrac{1}{2}u_{n+1}=u_{n+1}+\dfrac{1}{2}u_n\end{matrix}\right.\)
\(v_{n+1}=u_{n+1}+\dfrac{1}{2}u_n\Rightarrow\left\{{}\begin{matrix}v_2=u_2+\dfrac{1}{2}u_1=b+\dfrac{1}{2}a\\v_{n+1}=v_n\end{matrix}\right.\)
\(\Rightarrow v_{n+1}=b+\dfrac{1}{2}a\Rightarrow u_{n+1}=b+\dfrac{1}{2}a-\dfrac{1}{2}u_n\)
\(\Leftrightarrow u_{n+1}-\left(\dfrac{1}{3}a+\dfrac{2}{3}b\right)=-\dfrac{1}{2}\left[u_n-\left(\dfrac{1}{3}a+\dfrac{2}{3}b\right)\right]\)
\(t_n=u_n-\left(\dfrac{1}{3}a+\dfrac{2}{3}b\right)\Rightarrow\left\{{}\begin{matrix}t_1=u_1-\dfrac{1}{3}a-\dfrac{2}{3}b=\dfrac{2}{3}\left(a-b\right)\\t_{n+1}=-\dfrac{1}{2}t_n\end{matrix}\right.\)
\(\Rightarrow t_n=\dfrac{2}{3}\left(a-b\right)\left(-\dfrac{1}{2}\right)^{n-1}\Rightarrow u_n=t_n+\dfrac{1}{3}a+\dfrac{2}{3}b=\dfrac{2}{3}\left(a-b\right)\left(-\dfrac{1}{2}\right)^{n-1}+\dfrac{1}{3}a+\dfrac{2}{3}b\)
\(\Rightarrow limun=\lim\limits\left[\dfrac{2}{3}\left(a-b\right)\left(-\dfrac{1}{2}\right)^{n-1}+\dfrac{1}{3}a+\dfrac{2}{3}b\right]=0\)
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\(lim\left[\dfrac{2}{3}\left(a-b\right)\left(-\dfrac{1}{2}\right)^{n-1}+\dfrac{1}{3}a+\dfrac{2}{3}b\right]=\dfrac{1}{3}a+\dfrac{2}{3}b\)
