\(u_n=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2n-1}+\dfrac{1}{2n+1}\)
\(=1-\dfrac{1}{2n+1}\)
\(=\dfrac{2n}{2n+1}\)
\(u_1=\dfrac{2\cdot1}{2\cdot1+1}=\dfrac{2}{3}\)
Với \(n\ne-\dfrac{1}{2},n\ge2\) ta xét hiệu:
\(u_{n+1}-u_n=\dfrac{2\left(n+1\right)}{2\left(n+1\right)-1}-\dfrac{2n}{2n+1}\)
\(=\dfrac{\left(2n+2\right)\left(2n+1\right)-2n\left(2n+2-1\right)}{\left(2n+1\right)\cdot\left(2\left(n+1\right)-1\right)}\)
\(=\dfrac{4n^2+2n+4n+2-4n^2-4n+2n}{4n^2-1}\)
\(=\dfrac{2}{4n^2-1}\)
\(\Rightarrow u_{n+1}=\dfrac{2}{4n^2-1}+u_n\)
Đây là cấp số cộng có công sai \(d=\dfrac{2}{4n^2-1}\)
Áp dụng công thức: \(u_n=u_1+\left(n-1\right)\cdot d\)
\(u_{2019}=\dfrac{2}{3}+\left(2019-1\right)\cdot\dfrac{2}{4\cdot2019^2-1}=0,7\)
\(u_{2020}=\dfrac{2}{3}+\left(2020-1\right)\cdot\dfrac{2}{4\cdot2020^2-1}=0,7\)
