sum of the series

Question

find the sum of 1+ i/3+(i/3)^2+ (i/3)^3+......=?

Answer 1

1 + i/3 + (i/3)² + ...

 

= 1 + i/3 - 1/9 - i/27 + 1/81 - ...

 

= (1-1/9+1/81----) + i(1/3 -1/27+1/243-...)

 

1-1/9+1/81-... is geometric with r = -1/9 and has sum 1/(1-(-1/9)) = 1/(10/9) = 9/10

 

1/3 - 1/27 + 1/243-... is geometric with r = -1/9 and has sum (1/3)/(1-(-1/9)) = (1/3)/(10/9) = 3/10

 

So, the given series has sum (9/10) + (3/10)i