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 On Tue, Jan 3, 2012 at 22:00, Gareth McCaughan <gareth.mccaughan@pobox.com>wrote:
On Wednesday 04 January 2012 01:17:49 Dan Asimov wrote:
A few years ago I read in some math journal that a certain sum of reciprocal integer powers has an unexpectedly simple sum -- I think it was 3/5.
But I can't find the article or a reference to this fact. So I don't know just which reciprocal integer powers are being summed (or whether 3/5 is right).
I don't think this is what you're after -- it's too easy and the sum isn't 3/5 or much like it -- but
sum {n>=2,k>=2} 1/n^k = sum {n>=2} sum {k>=2} 1/n^k = sum {n>=2} 1/n(n-1) = sum {n>=2} 1/(n-1) - 1/n = 1
which is kinda cute. (Query: is there a "trivial" proof that regards each 1/n^k term as a probability or something and thereby makes it instantly obvious that the sum is 1?)
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