1. Tìm lim un
a. \(u_n=\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\)
b. \(u_n=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\)
c.\(u_n=\dfrac{1}{1}+\dfrac{1}{1+2}+...+\dfrac{1}{1+2+...+n}\)
d. \(u_n=\left[\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{n^2}\right)\right]\)
Ai giúp mk với hoặc gợi ý cho mik cx đc . Tks nhiều
a.
\(u_n=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(n-2\right)n}+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow\lim u_n=\lim\left(\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\right)=\dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}\)
b.
\(u_n=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}\)
\(\Rightarrow\lim u_n=\lim\left(1-\dfrac{1}{n+1}\right)=1\)
c.
\(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(\Rightarrow\dfrac{1}{1+2+...+n}=\dfrac{2}{n\left(n+1\right)}=\dfrac{2}{n}-\dfrac{2}{n+1}\)
\(\Rightarrow u_n=1+\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+...+\dfrac{2}{n}-\dfrac{2}{n+1}\)
\(=1+1-\dfrac{2}{n+1}=2-\dfrac{2}{n+1}\)
\(\Rightarrow\lim u_n=\lim\left(2-\dfrac{2}{n+1}\right)=2\)
d.
\(u_n=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{n^2}\right)\)
\(=\dfrac{\left(2^2-1\right)\left(3^2-1\right)...\left(n^2-1\right)}{2^2.3^2...n^2}\)
\(=\dfrac{1.3.2.4...\left(n-1\right)\left(n+1\right)}{2^2.3^2...n^2}\)
\(=\dfrac{1.2...\left(n-1\right)}{2.3...n}.\dfrac{3.4...\left(n+1\right)}{2.3...n}\)
\(=\dfrac{1}{n}.\dfrac{n+1}{2}=\dfrac{n+1}{2n}\)
\(\Rightarrow\lim u_n=\lim\left(\dfrac{n+1}{2n}\right)=\lim\left(\dfrac{1+\dfrac{1}{n}}{2}\right)=\dfrac{1}{2}\)
