Find the series?

Question

Find the series of (1/3)^i from i=n to infinity.

 

Answer: (3/2)(1/3)^n

 

 

Answer 1

S = sum{i=n,∞}((1/3)^i)

 

S = (1/3)^n + (1/3)^(n+1) + (1/3)^(n+2) + (1/3)^(n+3) + ...

 

(1/3) S = (1/3)^(n+1) + (1/3)^(n+2) + (1/3)^(n+3) + ...

 

S - (1/3) S = (1/3)^n

 

S = (1/3)^n/(1-1/3)

 

S = (1/3)^n/(2/3)

 

S = (3/2) (1/3)^n