Question

Cho \(S_n=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\). Kết quả giới hạn của limS_n bằng:

A. \(\dfrac{1}{3}\)

B. \(\dfrac{1}{5}\)

C. \(\dfrac{1}{2}\)

D. 0

Answer 1

Chọn B

Answer 2

B

Answer 3

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

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

-> chọn D 

Answer 4

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

\(\Rightarrow\lim S_n=\lim\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)=\dfrac{1}{2}\left(1-0\right)=\dfrac{1}{2}\)