@py
Bài làm :
Ta đặt :
S = 1 - $\frac{1}{2}$ - $\frac{1}{2^{2}}$ - ... - $\frac{1}{2^{10}}$
2S = 2 - 1 - $\frac{1}{2}$ - ... - $\frac{1}{2^{9}}$
2S - S = ( 2 - 1 - $\frac{1}{2}$ - ... - $\frac{1}{2^{9}}$ ) - ( 1 - $\frac{1}{2}$ - $\frac{1}{2^{2}}$ - ... - $\frac{1}{2^{10}}$ )
S = 2 - 1 - $\frac{1}{2}$ - ... - $\frac{1}{2^{9}}$ - 1 - $\frac{1}{2}$ - $\frac{1}{2^{2}}$ - ... - $\frac{1}{2^{10}}$
S = 1 - $\frac{1}{2}$ - ... - $\frac{1}{2^{9}}$ - 1 - $\frac{1}{2}$ - $\frac{1}{2^{2}}$ - ... - $\frac{1}{2^{10}}$
⇒ S = $\frac{1}{2^{10}}$ > $\frac{1}{2^{11}}$
⇒ S > $\frac{1}{2^{11}}$ → đpcm
