# 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+.....

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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)`