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

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Answer 1 · 2017-03-03T00:04:28

$R=1/e$

Use the ratio test, which states that $suma_n$ converges if $L<1$ , where $L=lim_(nrarroo)abs(a_(n+1)/a_n)$ . Here,

$L=lim_(nrarroo)abs((n+1)^(n+1)/((n+1)!)x^(n+1)(n!)/n^n1/x^n)$

Simplifying:

$L=lim_(nrarroo)abs((n+1)^(n+1)/n^n((n!)/((n+1)n!))x^(n+1)/x^n)$

$L=lim_(nrarroo)abs((n+1)^(n+1)/(n+1)1/n^nx)$

Bringing the $x$ out of the limit, since the limit depends only on how $n$ changes:

$L=absxlim_(nrarroo)abs((n+1)^n/n^n)$

$L=absxlim_(nrarroo)abs(((n+1)/n)^n)$

This is a well-known limit that approaches $e$ :

$L=eabsx$

The posted series will converge when $L<1$ , so the interval of convergence will be on when:

$eabsx<1$

$absx<1/e$

Thus the radius of convergence is $R=1/e$ , centered at $x=0$ .

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