What is the convergence radius for the series $sum_(r=0)^oo((x+1)/3)^r$ ?

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Answer 1 · 2016-11-20T15:34:29

$abs(x+1)<3$

You can write this series as

$sum_(r=0)^oo((x+1)/3)^r$ Calling now $y = (x+1)/3$ we have equivalently

$lim_(n->oo)sum_(r=0)^ny^r=lim_(n->oo)(y^n-1)/(y-1)$ .

We are using the fact

$1+y+y^2+cdots+y^n=(y^(n+1)-1)/(y-1)$

Now, if $absy < 1$ we have $lim_(n->oo)y^(n+1)=0$ so assuming $absy < 1$

$lim_(n->oo)sum_(r=0)^ny^r = 1/(1-y) = 1/(1-(x+1)/3) = 3/(2-x)$ this of course for $abs(x+1) < 3$

What is the convergence radius for the series sum_(r=0)^oo((x+1)/3)^rsum_(r=0)^oo((x+1)/3)^r ?