Question

Consider two infinite geometric series. The first has leading term $a,$ common ratio $b,$ and sum $S.$ The second has a leading term $b,$ common ratio $a,$ and sum $1/S.$ Find the value of $a+b.$

Answer 1

a ab ab^2 ab^3 .....
infinite so sum = a/ (1-b) = s

b ba ba^2 ba^3 ....
sum = b/(1-a) = 1/s

a = s -bs
b = 1/s -a/s
a+b = s - b s + 1/s -a/s

Answer 2

as a number?

Answer 3

well, a+b = s + 1/s - (a/s+ bs)
I think that only works if s = 1
a+b = 2 - (a+b)
2(a+b) =2
a+b =1

Answer 4

How about this:
From Damons:
a/(1-b) = s and
b/(1-a) = 1/s <b>OR</b> (1-a)/b = s

so a/(1-b) = (1-a)/b
ab = 1 - a - b + ab
a+b = 1

Answer 5

Thanks bro