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find radius of convergence by explaining clearly
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\[\sum_{n=0}^{\infty}x^n\]
x=1 must diverge because the limit as n goes to infinity is 1 which isnt 0 x>1 must diverge because the limit as n goes to infinity is infinity which isnt 0 x=-1 must diverge because the limit as n goes to infinity is -1 or 1 which means it doesnt exist and that isnt equal to 0 x<-1 must diverge because the limit as n goes to infinity doesnt exist so u have -1<x<1 to check now. I kno 0<x<1 converges as the geometric series so -1<x<0 must converge because it converges absolutely as the geometric series \[\left| (-x)^n \right|=x^n\]
so diameter is 1-(-1)=2 so radius is 1
jahtoday ,great answer thanks for ur clear explaining
no prob
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