what is the sum [n= 0 -> infinity] of the the series x^2/(1 +x^2)^n for x not equal to 0?

Answer 1

This is a geometric series with first term (a)
equal to x^2
and common ratio (r)
equal to 1/(1 + x^2)

Hence the limiting sum is
a/(1-r)
= (x^2)/[1 - 1/(1+ x^2)]

Multiply top and bottom by 1 + x^2, and do some simplifying, including cancelling out the common factor x^2 from top and bottom, and I think you should get
1+ x^2 as the answer.

Answer 2

This is a geometric series, so the sum is the first term divided by(1 minus the common ratio): The first term is x^2/1=x^2

The ratio is 1/(1+x^2). Thus the sum is

x^2/(1-1/(1+x^2)) = x^2/(x^2/(1+x^2)) = 1+x^2

Answer 3

If you pull x^2 out of the sum the rest is geometric series whose sum = (x^2+1)/x^2. Therefore your sum = x^2+1.