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)#