Question

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

Answer 1

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]}