Question

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

Answer 1

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

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

Explanation:

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`