How do you find the radius of convergence $Sigma x^(6n)/(n!)$ from $n=[1,oo)$ ?

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Answer 1 · 2017-10-27T14:16:56

The answer is $-oo < x < +oo$

Apply the ratio test to compute the convergence interval

$lim_(n->oo)|((x^(6(n+1))/((n+1)!))/(x^(6n)/(n!)))|$

$=lim_(n->oo)(|x^6/(n+1)|)$

$=|x^6|lim_(n->oo)(|1/(n+1)|)$

$=|x^6|*0$

$=0$

As the limit is $<1$ for every $x$ , therefore $sum_(n=1)^(+oo)x^(6n)/(n!)$ converges for all $x$

How do you find the radius of convergence Sigma x^(6n)/(n!)Sigma x^(6n)/(n!) from n=[1,oo)n=[1,oo)?