How do you find the radius of convergence $Sigma n^nx^n$ from $n=[1,oo)$ ?

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Answer 1 · 2017-01-18T09:04:44

The radius of convergence is $R=0$ , as the series

$sum_(n=1)^oo n^nx^n$

is divergent for any $x !=0$

Given the series:

$sum_(n=1)^oo n^nx^n$

we can apply the ratio test:

$abs (a_(n+1)/a_n) = (n+1)^(n+1)/n^n abs(x)^(n+1)/abs(x)^n = (n+1)((n+1)/n)^nabs(x)$

so that:

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

as:

$lim_(n->oo) (1+1/n)^n = e$

for any $x>0$ we have:

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

and the series is divergent, so the radius of convergence is $R=0$

How do you find the radius of convergence Sigma n^nx^nSigma n^nx^n from n=[1,oo)n=[1,oo)?