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

1 Answer
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#