5

I am aware of the following formula: $$\sum_{n=0}^{\infty}(-1)^nr^n=\frac{1}{1+r}$$

However, I am having difficulty understanding if there is a simple formula for the following equation:

$$\sum_{n=0}^{\infty}(-1)^nr^{2n}={ }?$$

In addition, what about the following?

$$\sum_{n=0}^{\infty}(-1)^{n+1}r^{2n+1}={ }?$$

I'm not sure how to go about identifying how to modify the original formula to fit the previous two, if that is even possible. Any help would be greatly appreciated.

Tyler Murry
  • 735
  • Can you use the formula $(x+1)^2=x^2+2x+1$ to expand $(x^2+1)^2$? If so, a similar operation on your first sum will give the second. – David Jul 03 '14 at 00:33
  • 1
    It may be best to think of $1+t+t^2+\cdots=\frac{1}{1-t}$ (when $|t|\lt 1$) as the basic formula, and use it to find all other related "sums." – André Nicolas Jul 03 '14 at 00:35

1 Answers1

9

$$\sum_{n=0}^{\infty}(-1)^nr^n=\sum_{n=0}^{\infty}(-r)^n=\cfrac{1}{1-(-r)}=\cfrac{1}{1+r}$$ So you can do the same for two following series : $$\sum_{n=0}^{\infty}(-1)^nr^{2n}=\sum_{n=0}^{\infty} {(-r^2)}^n $$$$\sum_{n=0}^{\infty}(-1)^{n+1}r^{2n+1}=-r\sum_{n=0}^{\infty}(-1)^nr^{2n}$$

Fabien
  • 2,958
  • @TylerMurry, in that case, the fact that the serie is an alternate one doesn't really matter since you can recognize a geometric serie (I assumed $|r|<1$). Moreover, $-r^2$ is negative. – Fabien Jul 03 '14 at 01:23