What is the radius of convergence of #sum_(n=1)^oo sin(x/n^2)#?

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Answer 1 · 2017-12-14T23:11:39

#sum_(n=0)^oo = sin(x/n^2)#

is absolutely convergent with radius of convergence #R=oo#

As for any #alpha in RR#:

#-alpha <= sin alpha <= alpha#

we have that:

#abs (sin (x/n^2)) <= abs (x/n^2)#

Now:

#sum_(n=0)^oo abs(x)/n^2 = abs(x) sum_(n=0)^oo 1/n^2 =(pi^2abs(x))/6#

is convergent for any #x in RR#, so by direct comparison also:

#sum_(n=0)^oo = sin(x/n^2)#

is absolutely convergent for any #x#, that is with radius of convergence #R=oo#