cho dãy số (un):\(\left\{{}\begin{matrix}u_1=3\\u_{n+1}=u_n^2-3u_n+4\end{matrix}\right.\)
Tìm lim\(\left(\dfrac{1}{u_1-1}+\dfrac{1}{u_2-1}+...+\dfrac{1}{u_n-1}\right)\)
Cho dãy un được xác định bởi
\(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=\dfrac{u_n}{u_n+1}\end{matrix}\right.\) với n=1,2,3,.... Tính
\(\lim\limits_{ }\dfrac{2014\left(u_1+1\right)\left(u_2+1\right)....\left(u_n+1\right)}{2015n}\)
\(u_{n+1}=\dfrac{u_n}{u_n+1}\Rightarrow\dfrac{1}{u_{n+1}}=\dfrac{1}{u_n}+1\)
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 \(d=1\Rightarrow v_n=v_1+\left(n-1\right).1=n\)
\(\Rightarrow u_n=\dfrac{1}{n}\)
\(\Rightarrow u_n+1=\dfrac{n+1}{n}\)
\(\lim\dfrac{2014\left(\dfrac{2}{1}\right)\left(\dfrac{3}{2}\right)\left(\dfrac{4}{3}\right)...\left(\dfrac{n+1}{n}\right)}{2015n}=\lim\dfrac{2014\left(n+1\right)}{2015n}=\dfrac{2014}{2015}\)
Cho dãy (un) \(\left\{{}\begin{matrix}u_1=\dfrac{1}{2}\\u_n=\dfrac{\sqrt{u_{n-1}^2+4u_{n-1}}+u_{n-1}}{2}\forall n\ge2\end{matrix}\right.\)
Tinh \(\lim\limits_{n\rightarrow+\infty}\left(\dfrac{1}{u_1^2}+\dfrac{1}{u_2^2}+...+\dfrac{1}{u_n^2}\right)\)
\(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=\dfrac{1}{3}\left(1+\dfrac{1}{u_n}\right)u_n\end{matrix}\right.\). gọi \(S_n=u_1+\dfrac{u_2}{2}+\dfrac{u_3}{3}+...+\dfrac{u_n}{n}\). tìm \(\lim\limits S_n\)
\(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=\dfrac{u_n^{2016}}{2015}+u_n\end{matrix}\right.\). Tính \(s=lim\left(\dfrac{u_1^{2015}}{u_2}+\dfrac{u_2^{2015}}{u_3}+...+\dfrac{u_n^{2015}}{u_{n+1}}\right)\)
Cho dãy số \(\left(U_n\right)\) được xác định bởi: \(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=\dfrac{1}{2}.\left(u_n+\dfrac{2}{u_n}\right)\end{matrix}\right.\), \(\forall n\ge1\). Tìm lim Un
Cho dãy số Un xác định bởi: \(\left\{{}\begin{matrix}u_1=\dfrac{1}{4}\\u_{n+1}=u_n^2+\dfrac{u_n}{2}\end{matrix}\right.\) với mọi \(n\ge1\). Tìm lim Un
Tính lim Un , biết :
a) \(\left\{{}\begin{matrix}U_1=\sqrt{2}\\U_{n+1}=\sqrt{2+U_n}\end{matrix}\right.\) , n \(\ge\) 1
b) \(\left\{{}\begin{matrix}U_1=\dfrac{1}{2}\\U_{n+1}=\dfrac{1}{2-U_n}\end{matrix}\right.\) .
Hiện tại mới nghĩ được câu b thôi
b/ \(u_1=\dfrac{1}{2};u_2=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3};u_3=\dfrac{1}{2-\dfrac{2}{3}}=\dfrac{3}{4}...\)
Nhận thấy \(u_n=\dfrac{n}{n+1}\) , ta sẽ chứng minh bằng phương pháp quy nạp
\(n=k\Rightarrow u_k=\dfrac{k}{k+1}\)
Chứng minh cũng đúng với \(\forall n=k+1\)
\(\Rightarrow u_{k+1}=\dfrac{k+1}{k+2}\)
Ta có: \(u_{k+1}=\dfrac{1}{2-u_k}=\dfrac{1}{2-\dfrac{k}{k+1}}=\dfrac{k+1}{k+2}\)
Vậy biểu thức đúng với \(\forall n\in N\left(n\ne0\right)\)
\(\Rightarrow limu_n=lim\dfrac{n}{n+1}=lim\dfrac{1}{1+\dfrac{1}{n}}=1\)
\(\left\{{}\begin{matrix}u_1=\dfrac{1}{2};u_2=3\\u_{n+2}=\dfrac{u_{n+1}.u_n+1}{u_{n+1}+u_n}\end{matrix}\right.\). tìm \(\left(u_n\right)\)
Tìm \(lim\dfrac{u_n}{3^n}\) biết: \(\left\{{}\begin{matrix}u_1=1\\u_{n+1}=3u_n+2n-1\end{matrix}\right.\)
\(u_n:\left\{{}\begin{matrix}u_1=1\\u_{n+1}=3u_n+2n-1\left(1\right)\end{matrix}\right.\)
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 \(lim\dfrac{1}{n}=0\rightarrow lim\dfrac{n}{3^n}=0\)
\(u_{n+1}=3u_n+2n-1\rightarrow v_{n+1}=3v_n\\ \rightarrow v_n=v_1.3^{n-1}=2.3^{n-1}\\ \rightarrow u_n=2.3^{n-1}-n\\ lim\dfrac{u_n}{3^n}=lim\dfrac{2.3^{n-1}-n}{3^n}=lim\left(\dfrac{2}{3}-\dfrac{n}{3^n}\right)=\dfrac{2}{3}\)