\(u_{n+1}=\dfrac{u_n}{u_n+1}\Rightarrow\dfrac{1}{u_{n+1}}=\dfrac{1}{u_n}+1\)

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 \(d=1\Rightarrow v_n=v_1+\left(n-1\right).1=n\)

\(\Rightarrow u_n=\dfrac{1}{n}\)

\(\Rightarrow u_n+1=\dfrac{n+1}{n}\)

\(\lim\dfrac{2014\left(\dfrac{2}{1}\right)\left(\dfrac{3}{2}\right)\left(\dfrac{4}{3}\right)...\left(\dfrac{n+1}{n}\right)}{2015n}=\lim\dfrac{2014\left(n+1\right)}{2015n}=\dfrac{2014}{2015}\)

Dễ dàng nhận thấy \(u_n\) là dãy dương

Ta sẽ chứng minh \(u_n< 2\) ; \(\forall n\)

Với \(n=1\Rightarrow u_1=\sqrt{2}< 2\) (thỏa mãn)

Giả sử điều đó đúng với \(n=k\) hay \(u_k< 2\)

Ta cần chứng minh \(u_{k+1}< 2\)

Thật vậy, \(u_{k+1}=\sqrt{u_k+2}< \sqrt{2+2}=2\) (đpcm)

Do đó dãy bị chặn trên bởi 2

Lại có: \(u_{n+1}-u_u=\sqrt{u_n+2}-u_n=\dfrac{u_n+2-u_n^2}{\sqrt{u_n+2}+u_n}=\dfrac{\left(u_n+1\right)\left(2-u_n\right)}{\sqrt{u_n+2}+u_n}>0\) (do \(u_n< 2\))

\(\Rightarrow u_{n+1}>u_n\Rightarrow\) dãy tăng

Dãy tăng và bị chặn trên nên có giới hạn hữu hạn. Gọi giới hạn đó là k>0

Lấy giới hạn 2 vế giả thiết:

\(\lim\left(u_{n+1}\right)=\lim\left(\sqrt{u_n+2}\right)\Leftrightarrow k=\sqrt{k+2}\)

\(\Leftrightarrow k^2-k-2=0\Rightarrow k=2\)

Vậy \(\lim\left(u_n\right)=2\)

Số xấu thế nhỉ?

\(u_n=v_n+\dfrac{\sqrt{5}-3}{2}\)

\(\Rightarrow v_{n+1}+\dfrac{\sqrt{5}-3}{2}=-\dfrac{1}{3+v_n+\dfrac{\sqrt{5}-3}{2}}\)

\(\Rightarrow\left\{{}\begin{matrix}v_1=u_1-\dfrac{\sqrt{5}-3}{2}=\dfrac{5-\sqrt{5}}{2}\\v_{n+1}=\dfrac{\dfrac{3-\sqrt{5}}{2}v_n}{\dfrac{3+\sqrt{5}}{2}+v_n}\end{matrix}\right.\)

\(v_n=\dfrac{1}{y_n}\Rightarrow\dfrac{1}{y_{n+1}}=\dfrac{\dfrac{3-\sqrt{5}}{2}.\dfrac{1}{y_n}}{\dfrac{3+\sqrt{5}}{2}+\dfrac{1}{y_n}}\)

\(\Rightarrow\dfrac{1}{y_{n+1}}=\dfrac{3-\sqrt{5}}{2y_n\left(\dfrac{3+\sqrt{5}}{2}+\dfrac{1}{y_n}\right)}=\dfrac{3-\sqrt{5}}{\left(3+\sqrt{5}\right)y_n+2}\)

\(\Leftrightarrow y_{n+1}=\dfrac{\left(3+\sqrt{5}\right)y_n}{3-\sqrt{5}}+\dfrac{2}{3-\sqrt{5}}\)

\(\Rightarrow\left\{{}\begin{matrix}y_1=\dfrac{1}{v_1}=\dfrac{2}{5-\sqrt{5}}\\y_{n+1}=\dfrac{3+\sqrt{5}}{3-\sqrt{5}}y_n+\dfrac{2}{3-\sqrt{5}}\end{matrix}\right.\)

\(z_n=y_n+\dfrac{\sqrt{5}}{5}\Rightarrow\left\{{}\begin{matrix}z_1=y_1+\dfrac{\sqrt{5}}{5}=\dfrac{5+3\sqrt{5}}{10}\\z_{n+1}=\dfrac{3+\sqrt{5}}{3-\sqrt{5}}z_n\end{matrix}\right.\)

\(\Rightarrow z_n:csn-co:\left\{{}\begin{matrix}z_1=\dfrac{5+3\sqrt{5}}{10}\\q=\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\end{matrix}\right.\)

\(\Rightarrow z_{n+1}=\dfrac{5+3\sqrt{5}}{10}.\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n\)

\(\Rightarrow y_{n+1}=z_{n+1}-\dfrac{\sqrt{5}}{5}=\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{\sqrt{5}}{5}\)

\(v_{n+1}=\dfrac{1}{y_{n+1}}=\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{\sqrt{5}}{5}}\)

\(u_{n+1}=v_{n+1}+\dfrac{\sqrt{5}-3}{2}=\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{\sqrt{5}}{5}}+\dfrac{\sqrt{5}-3}{2}\)

Xét: 

\(u_{n+2}-u_{n+1}=\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}.\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n.\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)-\dfrac{\sqrt{5}}{5}}+\dfrac{\sqrt{5}-2}{2}-\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{\sqrt{5}}{5}}-\dfrac{\sqrt{5}-2}{2}\)

\(=\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n.\dfrac{3+\sqrt{5}}{3-\sqrt{5}}-\dfrac{\sqrt{5}}{5}}-\dfrac{1}{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{\sqrt{5}}{5}}\)

\(=\dfrac{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n-\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n.\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)}{.....}\)

\(=\dfrac{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n\left(1-\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)}{....}=\dfrac{\dfrac{5+3\sqrt{5}}{10}\left(\dfrac{3+\sqrt{5}}{3-\sqrt{5}}\right)^n.\left(-\dfrac{5+3\sqrt{5}}{2}\right)}{...}< 0\)

\(\Rightarrow\) dãy giảm

\(\Rightarrow u_1>u_2>....>u_n\)

\(\Rightarrow\lim\limits u_n=1\)

Bn tham khảo đây nhé: https://diendantoanhoc.org/topic/140204-t%C3%A0i-li%E1%BB%87u-d%C3%A3y-s%E1%BB%91/

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 \(\forall n\in N\left(n\ne0\right)\)

\(\Rightarrow limu_n=lim\dfrac{n}{n+1}=lim\dfrac{1}{1+\dfrac{1}{n}}=1\)