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 $R=oo$

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