Đặt \(\frac{1}{a}\)+\(\frac{1}{a^2}\)+...+\(\frac{1}{a^n}\)=A
A.a= 1+\(\frac{1}{a}\)+...+\(\frac{1}{a^{n-1}}\)
A.(a-1)=1+\(\frac{1}{a}\)+...+\(\frac{1}{a^{n-1}}\)- \(\frac{1}{a}\)+\(\frac{1}{a^2}\)+...+\(\frac{1}{a^n}\)
A.(a-1)=1- \(\frac{1}{a^n}\)
A.(a-1)<1
A<\(\frac{1}{a-1}\)
Vậy \(\frac{1}{a}\)+\(\frac{1}{a^2}\)+...+\(\frac{1}{a^n}\)<\(\frac{1}{a-1}\)