How to graph \sum_{n=0}^\oo 2x^n?

1 Answer
Jan 1, 2018

Suppose that sum_(n=0)^oo2x^n=S.

First, this series is only convergent when -1< x<1, as an infinite sum which summation terms does not tend to 0 must be divergent. So, lim_(n->oo)x^n exists if and only if -1< x<1

Then,
2+sum_(n=1)^oo2x^n=S
2+xsum_(n=1)^oo2x^(n-1)=S
2+xsum_(n=0)^oo2x^n=S

Since S=sum_(n=0)^oo2x^n,
2+xS=S

Solving for S, we obtain S=2/(1-x), which is only valid when -1< x<1 (else the initial infinite sum is undefined).

With this, we can easily graph the value of sum_(n=0)^oo2x^n for different values of x:
graph{2/(1-x)sqrt(-(x+1)(x-1))/sqrt(-(x+1)(x-1)) [-2, 2, -1, 10]}