Proving simplification of summation

Question

I'm trying to prove that $\frac{1}{1+z^2}=1-z^2+z^4-z^6+...$ for $|z|<1$. The only thing that I can think of is that $\sum_{n=0}^\infty(-1)^nz^{2n}=1-z^2+z^4-z^6+...$, but I'm rather lost. Can anyone give me a hint please?

Answer 1

$$\sum_{n=0}^{\infty}(-1)^nz^{2n}$$ is the correct way to write it. But the next thing you should note is the geometric series formula: $$\sum_{n=0}^{\infty}z^n=\frac1{1-z}$$ If you substitute $(iz)^2$ into the second equation, you reach $$\sum_{n=0}^{\infty}(iz)^{2n}=\frac1{1-(iz)^2}$$ And since $i^2=-1$, we have $$\sum_{n=0}^{\infty}(-1)^nz^{2n}=\frac1{1+z^2}$$ QED

Answer 2

Hint: $(1-y)\sum_{j=0}^ny^j=1-y^{n+1}.$

Let $y=-z^2.$