Find the sum of the series when it converges

Question

1+2r+r^2+2r^3+r^4+2r^5+r^6...

 

We we only know how to work with basic geometric series at this point. We are told it converges for |r|<1 so I thought a/(1-r) applies and a is 1. 

I don't see a constant ratio, so it doesn't make sense to me. What am I missing?

Answer 1

1+2r+r²+2r³+r⁴+... = (1+r²+r⁴+r⁶+...) + (2r+2r³+2r⁵+...)

 

                                    = (1+r²+r⁴+r⁶+...) + 2r(1+r²+r⁴+r⁶+...)

 

                                    = (1+2r)(1+r²+r⁴+r⁶+...)

 

1+r²+r⁴+r⁶+... is geometric with common ratio r².  The series converges as long as -1<r²<1, which is true if -1<r<1.  So, assuming that -1<r<1, the given series has sum:

 

        (1+2r)[1/(1-r²)]  = (1+2r)/(1-r²)