use the properties of the geometric series to find the sum of the series. for what values of the variable does the series converge to this sum?y-y^2+y^3-y^4+.....
Opps, I made a mistake
`sum_(n=0)^oo ar^k = a/(1-r)`
we have `y - y^2 + y^3 - y^4 ...`
The real answer is `a = y, r = -y.` This is correct and we get
`y(-y)^0 + y(-y)^1 + y(-y)^2 + ...` this gives us the above series.
So our answer is `y/(1-(-y)) = y/(1+y)`
`sum_(n=1)^oo ar^k = a/(1-r)`
In this case r = -y and a = 1. So the sum of this series is
`1/(1-(-y)) = 1/(1+y)`