Prove given properties of infinite sums

Proof. (Not sure if I’m missing something deeper here…) Since we are given that
$1 + x^2 + x^4 + \cdots + x^{2n} + \cdots = \frac{1}{1-x^2} \qquad \text{for } |x| < 1$

we have

$1 + 0 + x^2 + 0 + x^4 + 0 + \cdots + x^{2n} + 0 + \cdots = \frac{1}{1-x^2} \qquad \text{for } |x| < 1. \qquad \blacksquare$
Proof. Given that
$1 + x + x^2 + \cdots + x^n + \cdots = \frac{1}{1-x} \qquad \text{for } |x| < 1$

we apply Theorem 10.2 to this and the result of part (a) so,

$\begin{align*} (1+x+x^2+\cdots + x^n + \cdots) - (1 + 0 + x^2 + 0 + x^4 + \cdots) &= \frac{1}{1-x} - \frac{1}{1-x^2}\\ \implies x + x^3 + x^5 + \cdots &= \frac{1-x^2-1+x}{(1-x)(1-x^2)} \\ &= \frac{-x(x-1)}{(1-x)(1-x^2)} \\ &= \frac{x}{1-x^2}. \qquad \blacksquare \end{align*}$
Proof. Applying Theorem 10.2 directly to the given expressions we have,
$\sum_{n=0}^{\infty} x^n - \sum_{n=0}^{\infty} x^{2n} = \sum_{n=0}^{\infty} (x^n - x^{2n}) = \frac{x}{1-x^2}. \qquad \blacksquare$

Stumbling Robot