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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.
\(u_n=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}\)
\(\Rightarrow\lim u_n=\lim\left(1-\dfrac{1}{n+1}\right)=1\)
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{99^2}\right)\left(1-\dfrac{1}{100^2}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{99}\right)\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}.\dfrac{3}{2}.\dfrac{2}{3}.\dfrac{4}{3}...\dfrac{98}{99}.\dfrac{100}{99}.\dfrac{99}{100}.\dfrac{101}{100}\)
\(=\dfrac{1.2...98.99}{2.3...99.100}.\dfrac{3.4...100.101}{2.3...99.100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)
1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+=???
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TL:
1+1+=2
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