Question

Cho tổng gồm 2019 số hạng 

S=\(\frac{1}{4}\)+ \(\frac{2}{4^2}\)+ \(\frac{3}{4^3}\)+......+\(\frac{2019}{4^{2019}}\)

chứng minh răng S<\(\frac{1}{2}\)

Answer 1

\(4S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}.\)

\(4S-S=3S=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2019}{4^{2018}}-\frac{1}{4}-\frac{2}{4^2}-...-\frac{2018}{4^{2018}}-\frac{2019}{4^{2019}}=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2018}}-\frac{2019}{4^{2019}}\)

\(3S< A=1+\frac{1}{4}+...+\frac{1}{4^{2018}}\)\(\Rightarrow3A=4A-A=4-\frac{1}{4^{2018}}< 4\)(sau khi rút gọn)

\(\Rightarrow3.3S< 4\Rightarrow9S< 4\)

\(\Rightarrow S< \frac{4}{9}< \frac{1}{2}\)