How do you find the interval of convergence $Sigma 2x^n$ from $n=[0,oo)$ ?

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Answer 1 · 2018-01-26T16:29:28

The radius of convergence is $abs(x)<1$ .

We will use the ratio test for convergence. We have that:

$sum_(n=0)^oo2x^n$

The ratio test tells us that the sum convergence if:

$lim_(n->oo)abs(a_(n+1)/a_n)<1$

In our case: $a_n=2x^n$

So:

$lim_(n->oo)abs(a_(n+1)/a^n)=lim_(n->oo)abs((2x^(n+1))/(2x^n))$

$=lim_(n->oo)abs(x^(n+1-1))=lim_(n->oo)abs(x)=abs(x)$

So, by the ratio test, for convergence:

$abs(x)<1$

How do you find the interval of convergence Sigma 2x^nSigma 2x^n from n=[0,oo)n=[0,oo)?