Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(u_n=\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\)
\(=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+...+\dfrac{1}{\left(n-1\right)\cdot\left(n+1\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+...+\dfrac{2}{\left(n-1\right)\left(n+1\right)}\right)\)
\(=\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{\left(n-1\right)}-\dfrac{1}{\left(n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{3}{4}-\dfrac{1}{2n+2}\)
\(\lim\limits u_n=\lim\limits\left(\dfrac{3}{4}-\dfrac{1}{2n+2}\right)\)
\(=\lim\limits\dfrac{3}{4}-\lim\limits\dfrac{1}{2n+2}\)
\(=\dfrac{3}{4}-\lim\limits\dfrac{\dfrac{1}{n}}{2+\dfrac{1}{n}}\)
=3/4
=>Chọn A