Sum of Infinite Geometric Sequence

Theorem
Let $x \in \R$ be a real number.

Let $\left|{x}\right| < 1$.

Then $\displaystyle \sum_{n=0}^\infty x^n$ is absolutely convergent to $\dfrac 1 {1 - x}$.

Corollary
With the same restriction on $x \in \R$:


 * $\displaystyle \sum_{n=1}^\infty x^n = \frac x {1-x}$

Proof
From Sum of Geometric Progression, we have:
 * $\displaystyle s_N = \sum_{n=0}^N x^n = \frac {1 - x^{N+1}} {1 - x}$

If $\left|{x}\right| < 1$, then by Power of a Number Less Than One $x^{N+1} \to 0$ as $N \to \infty$.

Hence $s_N \to \dfrac 1 {1 - x}$ as $N \to \infty$.

The result follows.

To demonstrate absolute convergence we note that if $x \ne 0$ and $\left|{x}\right| < 1$, it follows that $0 < \left|{x}\right| < 1$.

Then:
 * $\displaystyle \sum_{n=0}^\infty \left|{x}\right|^n = \frac 1 {1-\left|{x}\right|}$

as $\left|{x}\right|$ fulfils the same condition for convergence as $x$.

Proof of Corollary
Or: