Question

Cho \(M=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\). So sánh M với 4/9

Answer 1

\(4.M=4.\left(\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+...+\frac{2014}{4^{2014}}\right)=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2014}{4^{2013}}\)

=> 4M - M = \(1+\left(\frac{2}{4}-\frac{1}{4}\right)+\left(\frac{3}{4^2}-\frac{2}{4^2}\right)+...+\left(\frac{2014}{4^{2013}}-\frac{2013}{4^{2013}}\right)-\frac{2014}{4^{2014}}\)

=> 3.M = \(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}-\frac{2014}{4^{2014}}\)

Tính \(N=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2013}}\)

=> \(4.N=4+1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2012}}\)

=> 4N - N = 4 - \(\frac{1}{4^{2013}}\)=> N = \(\frac{4}{3}-\frac{1}{3.4^{2013}}\)=> N < 4/3

Ta có:  3M < N => M < N/3 => M < (4/3)/3 = 2/9

vậy M < 4/9