Question

tìm gioi hạn \(lim\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{2^{n-1}}\right)\)

Answer 1

\(S=1+\dfrac{1}{2}+...+\dfrac{1}{2^{n-1}}\)  là tổng cấp số nhân với \(\left\{{}\begin{matrix}u_1=1\\q=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow S=1.\dfrac{1-\left(\dfrac{1}{2}\right)^n}{1-\dfrac{1}{2}}=2-\dfrac{1}{2^{n-1}}\)

Do đó: \(\lim\left(S\right)=\lim\left(2-\dfrac{1}{2^{n-1}}\right)=2\)