Geometric series
"\\displaystyle\\sum_{i=0}^\\infin ar^n=\\dfrac{a}{1-r}, |r|<1" We have "a=1, r=\\dfrac{1}{3}, |r|=\\dfrac{1}{3}<1." Then
"\\displaystyle\\sum_{i=0}^\\infin (\\dfrac{1}{3})^n=1+\\dfrac{1}{3}+\\dfrac{1}{9}+\\dfrac{1}{27}+...=\\dfrac{1}{1-\\dfrac{1}{3}}=\\dfrac{3}{2}" Therefore
"1+3+\\dfrac{1}{3}+\\dfrac{1}{9}+\\dfrac{1}{27}+...=3+\\dfrac{3}{2}="
"=\\dfrac{9}{2}=4.5"
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