Giải thích các bước giải:
Ta có:
$S=\dfrac{1}{a^2}+\dfrac{2}{a^3}+\dfrac{3}{a^4}+...+\dfrac{n}{a^{n+1}}$
$\to aS=\dfrac{1}{a}+\dfrac{2}{a^2}+\dfrac{3}{a^3}+...+\dfrac{n}{a^{n}}$
$\to aS-S=\dfrac{1}{a}+\dfrac{1}{a^2}+\dfrac{1}{a^3}+...+\dfrac{1}{a^n}-\dfrac{n}{a^{n+1}}$
$\to (a-1)S=\dfrac{1}{a}+\dfrac{1}{a^2}+\dfrac{1}{a^3}+...+\dfrac{1}{a^n}-\dfrac{n}{a^{n+1}}$
$\to a(a-1)S=\dfrac{1}{1}+\dfrac{1}{a}+\dfrac{1}{a^2}+...+\dfrac{1}{a^{n-1}}-\dfrac{n}{a^{n}}$
$\to a(a-1)S-(a-1)S=\dfrac{1}{1}-\dfrac{n}{a^{n}}-\dfrac{1}{a^n}+\dfrac{n}{a^{n+1}}$
$\to (a-1)(a-1)S=1-\dfrac{n+1}{a^n}+\dfrac{n}{a^{n+1}}$
$\to (a-1)^2S=1+\dfrac{n}{a^{n+1}}-\dfrac{n+1}{a^n}$
Ta có $a^{n+1}>a^n\to \dfrac{n}{a^{n+1}}<\dfrac{n}{a^n}<\dfrac{n+1}{a^n}$
$\to \dfrac{n}{a^{n+1}}-\dfrac{n+1}{a^n}<0$
$\to 1+\dfrac{n}{a^{n+1}}-\dfrac{n+1}{a^n}<1$
$\to (a-1)^2S<1$
$\to S<\dfrac{1}{(a-1)^2}$
