+118658 for the series
infinity
E 5/10 ^n
n -> 1
find the partial sum and the sum
\(\displaystyle\sum_{n\rightarrow 1}^{\infty}\;\frac{5}{10^n}\\ \frac{5}{10}+\frac{5}{10^2}+\frac{5}{10^3}+\dots+\frac{5}{10^n}\\ \mbox{This is a GP a=0.5, r=0.1, n=n}\\ =0.5\times\frac{1-0.1^n}{1-0.1} \\ =0.5\times\frac{1-0.1^n}{0.9} \\ =\frac{5(1-0.1^n)}{9} \\~\\ \displaystyle\sum_{n\rightarrow 1}^{\infty}\;\frac{5}{10^n}=\frac{0.5}{1-0.1}=\frac{5}{9} \)
