\(S=5+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{3^n}+...\\\text{Xét tổng}\quad\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{3^n}+...\quad\text{là cấp số nhân lùi vô hạn với} u_1=\dfrac{1}{3};q=\dfrac{1}{3}\\\Leftrightarrow \dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{3^n}+...=\dfrac{u_1}{1-q}=\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}=\dfrac{1}{2}\\S=5+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{3^n}\\\Leftrightarrow 5+\dfrac{1}{2}=5\dfrac{1}{2}.\\\text{Vậy}\quad S=5\dfrac{1}{2}.\)
