How do you find the interval of convergence $Sigma (x^(2n)/(n!))$ from $n=[0,oo)$ ?

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Answer 1 · 2017-02-08T23:52:27

The series is absolutely convergent for every $x in RR$ and its sum is:

$sum_(n=0)^oo x^(2n)/(n!) = e^(x^2)$

Use the ratio test stating that a series:

$sum_(n=0)^oo a_n$

is absolutely convergent if:

$L = lim_(n->oo) abs(a_(n+1)/a_n) < 1$

and divergent for $L>1$

Calculate the ratio:

$abs(a_(n+1)/a_n) =abs ( (x^(2(n+1)) / ((n+1)!) )/ (x^(2n)/(n!))) = abs(x^(2n+2)/x^(2n)) (n!)/((n+1)!) =x^2/(n+1)$

Thus:

$lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) x^2/(n+1) = 0$

for any $x in RR$ .

We can also note that:

$sum_(n=0)^oo x^(2n)/(n!) = sum_(n=0)^oo (x^2)^n/(n!) = e^(x^2)$

How do you find the interval of convergence Sigma (x^(2n)/(n!))Sigma (x^(2n)/(n!)) from n=[0,oo)n=[0,oo)?