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#