Question

Anyone who knows this one?
Find the sum of Sn of the final geometric series:

1+1/3+1/3^2+.....+1/3^(n-1)

what is Sn approaching against when n goes towards infinity?

Answer 1

\[ ~~~~~~~~~~~S = 1 + \frac 13+\frac 1{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}\\~~~-\frac13 S = ~~-\frac 13-\frac 1{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{n-1}}-\frac{1}{3^n}\\----------------------\\(1-\frac13)S=1+~0+~~0+~~~0+...+~~~~~~0~~~-\frac 1{3^n}\]

Answer 2

\[(1-\frac13)S=1-\frac{1}{3^n}\]

\[S=\cfrac{1-\frac{1}{3^n}}{1-\frac13}\]
now, as n to infinity, 1/3^n to 0 leaving us with ....\[S=\frac{1}{1-\frac13}\]