Question

Cho \(S_1=1+\frac{1}{5}\), \(S_2=1+\frac{1}{5}+\frac{1}{5^2}\), \(S_3=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}\), tới \(S_n=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...........+\frac{1}{5^n}\). Chứng minh rằng : \(A=\frac{1}{5S_1^2}+\frac{1}{5^2S_2^2}+\frac{1}{5^3S_3^2}+\frac{1}{5^4S_4^2}+..........+\frac{1}{5^nS_n^2}<\frac{35}{36}\)

Answer 1

Khi \(n=1\to A=\frac{1}{5S_1^2}=\frac{5}{36}S_{k-1}\to S^2_k>S_k\cdot S_{k-1}\).

Vậy ta có \(\frac{1}{5^kS_k^2}