Question

What is the radius of convergence of `sum_1^oox/n^2`?

Answer 1

Infinite

Explanation:

The `x` can be moved outside the sum, which is a convergent sum.

`sum_(n=1)^oo x/n^2 = x sum_(n=1)^oo 1/n^2 = x*pi^2/6`

Proving that `sum_(n=1)^oo 1/n^2 = pi^2/6` is not easy, but showing that it converges is:

`sum_(n=1)^N 1/n^2 < 1+sum_(n=1)^N 1/(n(n+1))`

`= 1+sum_(n=1)^N (1/n-1/(n+1))`

`= 1+sum_(n=1)^N 1/n - sum_(n=1)^N 1/(n+1)`

`= 1+sum_(n=1)^N 1/n - sum_(n=2)^(N+1) 1/n`

`= 2 + color(red)(cancel(color(black)(sum_(n=2)^N 1/n))) - color(red)(cancel(color(black)(sum_(n=2)^N 1/n))) - 1/(N+1)`

`= 2-1/(N+1) -> 2` as `N->oo`