How do you find the radius of convergence $Sigma (n^nx^n)/(lnn)^n$ from $n=[2,oo)$ ?

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How do you find the radius of convergence $Sigma (n^nx^n)/(lnn)^n$ from $n=[2,oo)$ ?

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Answer 1 · 2017-03-01T00:45:13

The root test states that the series $suma_n$ is convergent if $L=lim_(nrarroo)root(n)abs(a_n)<1$ .

Here we see that $a_n=((nx)/lnn)^n$ , so:

$L=lim_(nrarroo)root(n)abs(((nx)/lnn)^n)=lim_(nrarroo)abs((xn)/lnn)$

Since the limit is dependent only on the change in $n$ , the $x$ can be removed from the limit like a constant:

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

We see that the limit approaches $oo$ , since $n$ grows faster than the logarithmic function $lnn$ . The only time when $L<1$ , when the series converges, will be when $L=0$ , which occurs only when $x=0$ .

Since there is only one value of $x$ for which the series converges, we see that $R=0$ .

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How do you find the radius of convergence $Sigma (n^nx^n)/(lnn)^n$ from $n=[2,oo)$ ?

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How do you find the radius of convergence Sigma (n^nx^n)/(lnn)^nSigma (n^nx^n)/(lnn)^n from n=[2,oo)n=[2,oo)?

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How do you find the radius of convergence $Sigma (n^nx^n)/(lnn)^n$ from $n=[2,oo)$ ?