Răspuns:
Explicație pas cu pas:
[tex]S=\dfrac{1}{2}+ \dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}+\dfrac{1}{1024}\\2*S=1+\dfrac{1}{2}+ \dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\\2*S-S=S=1+\dfrac{1}{2}+ \dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}-(\dfrac{1}{2}+ \dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}+\dfrac{1}{1024})=1-\dfrac{1}{1024}=\dfrac{1024}{1024}-\dfrac{1}{1024}=\dfrac{1023}{1024} <1[/tex]
