What is the interval of convergence of $sum_1^oo (x^n *n^n)/(n!)$ ?

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Answer 1 · 2015-11-05T19:18:14

$(-1/e, 1/e)$ or possibly $[-1/e, 1/e]$

Let $a_n = n^n/(n!)$

Then:

$a_(n+1) = (n+1)^(n+1)/((n+1)!) = (n+1)^n/(n!)=((n+1)/n)^n n^n/(n!)$

$=(1+1/n)^n a_n$

Now $(1+1/n)^n -> e$ as $n->oo$

So: $(x^(n+1)*(n+1)^(n+1))/((n+1)!) -: (x^n*n^n) / (n!) = (a_(n+1)/a_n)x -> ex$ as $n->oo$

Hence if $abs(x) < 1/e$ then $sum_(n=0)^oo (x^n*n^n) / (n!)$ is absolutely convergent.

If $abs(x) > 1/e$ then $EE N in ZZ$ such that:

$abs((x^(n+1)*(n+1)^(n+1))/((n+1)!) -: (x^n*n^n) / (n!)) > 1$ for all $n >= N$

so the sum diverges.

I am not sure about the case $abs(x) = 1/e$ , since $(1+1/n)^n < e$ for all $n in NN$ .

What is the interval of convergence of sum_1^oo (x^n *n^n)/(n!)sum_1^oo (x^n *n^n)/(n!)?