Geometric Sequence convergance without Taylor

So lately I have observed the following video:

The nice fellow on this video explained why this following property is correct $$\frac{1}{1-x}=1+x+x^2+x^3+x^4...=\sum_{n=0}^{\infty}x^n$$ as long as $|x|<1$

I did not quite understand two things:

1.the long devisioning he did in order to get the power seirie from the fraction of one over X minus one

2.why is it true only if $|x|<1$, why does this condition apply?

• If $|x|\ge1$ then $|x^n|\not\to0$ so the series diverges. – Lord Shark the Unknown Aug 10 '17 at 15:11

The video is not very rigorous, which is why you aren't able to determine why $|x|<1$ is needed. The series $\sum_{n=0}^\infty x^n$ really just means $$\lim_{n\to\infty}(1+x+x^2+\cdots+x^n).$$ Clearly this limit does not exist if $x=1$. Suppose $x\ne 1$. We attempt to show that this limit is equal to $1/(x-1)$ by showing that $$\lim_{n\to\infty}\left| (1+x+\cdots+x^n)-\frac{1}{1-x} \right|=0.$$ Note that for any $n$, we have $$\left| (1+x+\cdots+x^n)-\frac{1}{1-x} \right| = \left|\frac{1}{1-x}\right||(1-x)(1+x+\cdots+x^n)-1|$$ $$= \left|\frac{1}{1-x}\right||(1+x+\cdots+x^n)-(x+x^2+\cdots+x^{n+1})-1|$$ $$= \left|\frac{1}{1-x}\right||-x^{n+1}| = \left|\frac{1}{1-x}\right||x|^{n+1}.$$ Now $|x|^{n+1}$ goes to $0$ as $n$ approaches infinty if and only if $|x|<1$. In conclusion $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$ if and only if $|x|<1$.

I do not like the video much, so I will give you hints to understand this property by your own in a very easy and nice way.

Let us set $S_N= \sum_{n=0}^N x^n$ for $N\in \mathbf N$ and lets consider $|x|<1$. Just follow my three steps.

Step 1: Proof $S_N (1-x) = 1-x^N$ (telescop sum).

Step 2: Divide by $1-x\neq 0.$

Step 3: Calculate the limit $N\to \infty$ (and do not forget the definition of a series).

The basic identity, true for any $x$ and positive integer $n$ is that $(1-x)\sum_{k=0}^{n-1} x^k =1-x^{n}$. This is readily proved by induction.

Therefore, if $x \ne 1$, $\sum_{k=0}^{n-1} x^k =\dfrac{1-x^{n}}{1-x} =\dfrac{1}{1-x}-\dfrac{x^{n}}{1-x}$ or $\dfrac{1}{1-x}-\sum_{k=0}^{n-1} x^k =\dfrac{x^{n}}{1-x}$.

I will now show that if $0 < x < 1$ then $x^n \to 0$ and if $x > 1$ then $x^n \to \infty$. The case $x < 0$ is handled by replacing $x$ by $-x$ and writing $\dfrac{1}{1+x}-\sum_{k=0}^{n-1} (-1)^kx^k =\dfrac{(-1)^nx^{n}}{1+x}$.

Note: What follows is not original.

The basic inequality used is Bernoulli's, which states that $(1+a)^n \ge 1+na$ if $a \ge 0$ and $n$ is a positive integer. This is readily proved by induction.

If $x > 1$ then $x = 1+y$ where $y > 0$. Therefore $x^n =(1+y)^n \ge 1+ny \gt ny$ and this $\to \infty$ as $n \to \infty$ for any $y > 0$.

If $0 < x < 1$, then $x = \dfrac1{1+y}$ where $y =\dfrac1{x}-1$ satisfies $y > 0$. Therefore $x^n =\dfrac1{(1+y)^n} \le \dfrac1{1+ny} \lt \dfrac1{ny} \to 0$ as $n \to \infty$.