Question

Tìm lim u_n biết \(u_n=\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\)

Answer 1

\(u_n=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

\(=\dfrac{1}{2}\left[\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n-1}\right)-\left(\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n+1}\right)\right]\)

\(=\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{2n+1}{n\left(n+1\right)}\right)=\dfrac{3n^2+n-1}{4n\left(n+1\right)}\)

\(\Rightarrow limu_n=lim\dfrac{3n^2+n-1}{4n^2+4n}=lim\dfrac{3+\dfrac{1}{n}+\dfrac{1}{n^2}}{4+\dfrac{4}{n}}=\dfrac{3}{4}\)

Answer 2

\(\dfrac{1}{n^2-1}=\dfrac{1}{2}\cdot\dfrac{2}{\left(n-1\right)\left(n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

Khi đó:

 \(u_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{3n^2+3n+2}{4n\left(n+1\right)}\)

\(lim_{u_n}=lim\dfrac{3n^2+3n+2}{4n\left(n+1\right)}=lim\dfrac{3+\dfrac{3}{n}+\dfrac{2}{n^2}}{4\left(1+\dfrac{1}{n}\right)}=\dfrac{3}{4}\)