Sum of an alternating geometric serie

Question

I have to find the sum of $$\sum_{n=0}^{\infty}\frac{(-1)^{n}}{s^{2n+2}}$$ I tried using geometric series, but I really don't know what to do with $(-1)^{n}$.

Since $$s^{2n+2}=(s^{2})^{n}\cdot s^{2}$$ I tried to write the serie as: $$\frac{1}{s^{2}}\sum_{n=0}^{\infty}(-s^{-2})^{n}$$ But I still don't know what to do.

Answer 1

the sum of a geometric series:

$\sum_\limits{n=0}^\infty a^n = \frac {1}{1-a}$

Now replace $a$ with $(-s^{-2})$

$(\frac 1{s^2})(\frac {1}{1-(-s^{-2})}) = \frac {1}{s^2 + 1}$

Answer 2

Hint: If $|u|<1$, one has $$\frac1{1+u}=1-u+u^2-u^3+\dotsm=\sum_{n\ge 0}(-1)^n u^n.$$