Giải thích các bước giải:
Ta có :
$1+\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^n}=\dfrac{\dfrac{1}{3^{n+1}}-1}{\dfrac{1}{3}-1}$
$\to 1+\dfrac{1}{3}+\dfrac{1}{3^2}+..+\dfrac{1}{3^n}=-\dfrac{1-3^{n+1}}{2\cdot \:3^n}$
$1+\dfrac{1}{2}+\dfrac{1}{2^2}+..+\dfrac{1}{2^n}=\dfrac{\dfrac{1}{2^{n+1}}-1}{\dfrac{1}{2}-1}$
$1+\dfrac{1}{2}+\dfrac{1}{2^2}+..+\dfrac{1}{2^n}=-\dfrac{1-2^{n+1}}{2^n}$
$\to\lim\dfrac{1+\dfrac{1}{3}+\dfrac{1}{9}+..+\dfrac{1}{3^n}}{1+\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{2^n}}$
$=\lim\dfrac{-\dfrac{1-3^{n+1}}{2\cdot \:3^n}}{-\dfrac{1-2^{n+1}}{2^n}}$
$=\lim\dfrac{2^{n}\left(1-3.3^{n}\right)}{2.3^n\left(1-2.2^{n}\right)}$
$=\lim\dfrac{2^{n}-3.6^{n}}{2.3^n-4.6^{n}}$
$=\lim\dfrac{(\dfrac{2}{6})^{n}-3}{2(\dfrac{3}{6})^n-4}$
$=\dfrac34$
