In my haste I left out 1/2 = sum {k>=2} 1/2^k. So the sum that adds up to 3/5 needs to include n=2 along with n>=11: sum {k>=2} 1/2^k + sum {n>=11,k>=2} 1/n^k = 3/5 - Robert On Wed, Jan 4, 2012 at 03:32, Robert Munafo <mrob27@gmail.com> wrote:
Gareth,
I think we can answer Dan's original question from your sum. Since
sum{k>=2} 1/n^k = 1/n sum{k>=1} 1/n^k = 1/n 1/(n-1) = 1/n(n-1),
and
1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90 = 2/5,
it follows that
1/110 + 1/132 + 1/156 + ... = sum {n>=11} 1/n(n-1) = sum {n>=11,k>=2} 1/n^k = sum {n>=2,k>=2} 1/n^k - sum {2<=n<=10,k>=2} 1/n^k = 1 - 2/5 = 3/5
(I hope I got all that right... :-)
- Robert
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com