How do you find the radius of convergence $Sigma (n^nx^n)/(ln(lnn))^n$ from $n=[3,oo)$ ?

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Answer 1 · 2017-03-02T21:58:36

The interval of convergence is the single value $x=0$ and the radius of convergence is $R=0$ .

$sum_(n=3)^oo(n^nx^n)/(ln(lnn))^n=sum_(n=3)^oo((nx)/ln(lnn))^n$

The root test says that the series $suma_n$ converges if $lim_(nrarroo)rootnabs(a_n)<1$ .

So, we will take $L=lim_(nrarroo)rootnabs(a_n)$ and find the values of $x$ that make $L<1$ , as this is when the series will converge.

$L=lim_(nrarroo)rootnabs(a_n)=lim_(nrarroo)rootnabs(((nx)/ln(lnn))^n)$

$color(white)L=lim_(nrarroo)abs((nx)/ln(lnn))$

Moving the $x$ from the limit, since the limit only depends on the changing of $n$ .

$L=absxlim_(nrarroo)abs(n/ln(lnn))$

Since $n$ grows faster than any logarithmic function, we see that $lim_(nrarroo)abs(n/ln(lnn))=oo$ .

The only time, then, when $L<1$ will be when the limit is multiplied by $0$ , that is, when $absx=0$ , which occurs only at $x=0$ .

Thus, the interval of convergence is the single value $x=0$ and the radius of convergence is $R=0$ .

How do you find the radius of convergence Sigma (n^nx^n)/(ln(lnn))^nSigma (n^nx^n)/(ln(lnn))^n from n=[3,oo)n=[3,oo)?