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

1 Answer
George C.
Nov 21, 2015

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

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