0
$\begingroup$

$\sum_{n=1}^{\infty} (1-p)^{2n + 1}$, assume $|1 - p| < 1$

I am unable to apply the formula $\sum_{n=1}^{\infty} x^n = \frac{1}{1 - x}$ as I dont have a $n$ power, what should I do?

  • 1
    $\sum_{n=1}^{\infty} x^n = \frac{x}{1 - x}$ – 2017-02-10
  • 0
    Hint: $x^{2n+1}=x\cdot(x^2)^n$. You do have an $n^{th}$ power. – 2017-02-10

1 Answers 1

0

$$\sum_{n=1}^{\infty} (1-p)^{2n + 1}=(1-p)\sum_{n=1}^{\infty} ((1-p)^{2})^n=\frac{(1-p)^3}{1-(1-p)^2}$$