Find the Sum

3/13 + 33/13^2 + 333/13^3 + ...... till infinity = 3[ 1/13 + 11/13^2 + 111/13^3 + ....till infinity ] = 3[ 1/13 + (1+10)/13^2 + (1+100)/13^3 + .....] = 3[ 1/13 + {1/13^2 + 10/13^2} + {1/13^3 + 100/13^3} +........infinity] = 3[ 1/13 + 1/13^2 + 10/13^2 + 1/13^3 + 100/13^3 +.] = 3[{1/13 + 1/13^2 + 1/13^3 +....} + { 10/13^2 + 100/13^3 =....}] Now we can split into sun of two expressions as : ONE : 3{1/13 + 1/13^2 + 1/13^3 +...... infinity} SECOND : 3{10/13^2 + 100/13^3 + 1000/13^4 +.... infi} Solutions are : for ONE : 3[(1/13)/{1-1/13}}] Formula for G.P. 3[(1/13)/(12/13)] for infinite series 3[1/12] 1/4 for SECOND : 3[10/13^2 + 100/13^3 + ....] Multiply with 10 and divide with 10 = (3/10)[100/13^2 + 1000/13^3 + ...] = (3/10)[10^2/13^2 + 10^3/13^3+....] adding 10/13 and substracting 10/13 in [] = (3/10)[10/13 + 10^2/13^2 + 10^3/13^3 +.... -10/13] applying infinite series formula = (3/10)[(10/13)/{1-10/13} - (10/13)] = (3/10)[(10/13)/(3/13) - (10/13)] = (3/10)[10/3 - 10/13] = 3[1/3 - 1/13] = 3[10/39] = 10/13 Now add the results of ONE and SECOND 1/4 + 10/13 = 53/52 So the series is resulted to the value : 53/52 Please respond for the correctness of my answer