How to show that $ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $? By long division, it is easy to show that $$ \frac{1}{1-x} = 1 + x + x^2 +x^3 +... $$
But how to show that
$$ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $$
 A: $$ \text{Let }S_{n}=1+x+x^2...+x^{n-1}$$
$$\implies xS_{n}=x+x^2....+x^{n}$$
Subtracting both equations,
$$S_{n}(1-x)=1-x^{n}$$
$$\implies S_{n}=\frac{1-x^n}{1-x}$$
Since it is an infinite series, $n\to\infty$ and it converges only when $|x|<1$. When $n\to\infty,x^n\to0$
$$\implies s_{\infty}=\frac{1-0}{1-x}=\frac{1}{1-x} $$
A: You have following expression-
$ 1 + x + x^2 +x^3 +...$
Now multiplying numerator and denominator by $x-1$, provided $x\neq1$, we have
$\frac{(1-x)( 1 + x + x^2 +x^3 +...)}{1-x}=\frac{(1+x+x^2...)-(x+x^2+x^3)}{1-x}=\frac{1}{1-x}$
A: Suppose $1+x+x^2+x^3...=S$ where $S \in R$.
If $x=0$ it is trivial, so suppose $x \neq 0$. Subtract each side by 1 and divide both sides by $x$.
This leaves: $S=(S-1)x^{-1}$. Solving for $x$ yields: $S*x=S-1$ which simplifies to $S(x-1)=-1$ which then yields $1+x+x^2+x^3...=(1-x)^{-1}$
Note domain restrictions of $|x|<1$ applies.
A: We start with the simple geometric series
$$\sum_{k=0}^n x^k = \dfrac{1-x^{n+1}}{1-x}$$
Now assume that $|x|<1$.
With this assumption in mind, taking the limit $n \to \infty$ gives,
$$\sum_{k=0}^{\infty} x^k = \dfrac{1}{1-x} \ , \ |x|<1$$
The upper $x$ term cancelled out due our assumption.
A: $$1+x+x^2+x^3+\dots$$
$$=(1+x)(1+x^2+x^4+\dots)$$
$$=(1+x)(1+x^2)(1+x^4+\dots)$$
$$=(1+x)(1+x^2)(1+x^4)\dots$$
$$=\frac{1-x^2}{1-x}\frac{1-x^4}{1-x^2}\frac{1-x^8}{1-x^4}\dots$$
$$=\frac{1}{1-x}$$
A: As this is an infinite $GP$ , and $|x|<1$ Then , $$S=\frac{1}{1-x}$$
