Sum of Infinite Geometric Sequence

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

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

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

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


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

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

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

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

The result follows.

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

Then $$\sum_{n=0}^\infty \left|{x}\right|^n$$ as $$\left|{x}\right|$$ fulfils the same condition for convergence as does $$x$$.

Proof of Corollary
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Or:

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