Consider the sum of the inverses of the squares of the primes; does it converge or diverge?

Well, sure.  Convergence of the sum of the inverses of the squares of the primes follows trivially from convergence of the sum of the inverses of the squares of the integers.  However, given the similar form of these two results, surely a satisfying proof must address the latter result?

Let S(n) = 1/1² + 1/2² + ... + 1/n²

Consider S(n) - 1 = 1/2² + 1/3² + ... + 1/n² < 1/2*1 + 1/3*2 + ... + 1/[n(n-1)]

Then we have 1/[n(n-1)] = 1/(n-1) - 1/n, so

S(n) - 1 < 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/(n-2) - 1/(n-1) + 1/(n-1) - 1/n, and so

S(n) - 1 < 1 - 1/n, and S(n) < 2 - 1/n

This shows that the sequence of partial sums, {S(n)}, which is strictly increasing, is bounded above (by 2), and therefore converges.

Edited on April 27, 2004, 4:22 pm

Posted by Nick Hobson on 2004-04-27 16:15:47