Question

[hoàng tử băng]

1/2+1/2^2+1/2^3+....+1/2^2013<1

Answer 1

[Xyz OLM]

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\)

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

Khi đó \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\right)\)

=> \(A=1-\frac{1}{2^{2013}}< 1\left(\text{Đpcm}\right)\)