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

1 Answer
NickTheTurtle
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]}

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