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

Answer 1

The sum is 1+ (1/x²)

To understand why, first do the sum ∑ (1/z^n) for n = 0 to ∞. This sum comes to z / (z-1). Then make the subsitution z = 1 + x².

If you don't understand the first part, let S be the sum ∑ (1/z^n)

S = 1 + 1/z + 1/z² + etc.
S/z = 1/z + 1/zx² + etc.
S(1 - 1/z) = 1
S = 1/(1-1/z) = z/(z-1)