Proving that if |x| < 1, then $\sum_{i=0}^{\infty}x^i=\frac{1}{1-x}$

Question

I'm supposed to use the answer from the previous question:

Prove inductively that for any real number $x \neq 1$, for any integer $n \geq 0$, we have $$1 + x + x^2 + \cdots + x^{n-1} + x^n = \frac{x^{n+1}−1}{x−1}$$

I got the answer to this, it's a rather easy simplification of the geometric sum by assuming the hypothesis is correct.

However, I'm not sure exactly how to prove the following using THAT result:

Prove using the answer above, that if |x| < 1, then $\sum_{i=0}^{\infty}x^{i}=\frac{1}{1-x}$

I'm completely lost.

Answer 1

Hint

$$\sum_{k=0}^\infty x^k:=\lim_{n\to\infty }\sum_{k=0}^nx^k.$$

Answer 2

Some big hints

Define:

$S_n(x) = \sum_{i=0}^{i=n} x^i$

Prove that

$x S_n(x) + 1 = S_{n+1}(x)$ Equation (1)

Next prove

$S_{n+1}(x) - S_n(x) = x^{n+1}$ Equation (2)

Finally substitue our definition of $S_{n+1}(x)$ from equation 1 into equation 2

Factorize so that $S_n(x)$ only appears once. Then rearrange the expression so it's of the form $S_n(x) = ...$