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?)

-- 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

As for your "trivial" proof, all I can come up with is a square divided into rectangles. Start with a 1x1 square and make marks along the bottom edge at points that are 1/2, 1/3, 1/4, 1/5, ... from the right-hand end (or 1/2, 2/3, 3/4, 4/5, ... from the left-hand end). Then draw vertical lines at these points to partition the square into rectangles whose widths (and therefore areas) are 1/n(n-1) for all n>=2. Then each rectangle can be partitioned into sub-rectangles of area 1/n^2 + 1/n^3 + 1/n^4 + ... The first rectangle, whose area is 1/2, should be divided in half to form 1/4 + 1/4. Then divide the upper half in half again to make 1/4+1/8+1/8. Divide one of these 1/8 rectangles in half to get 1/4+1/8+1/16+1/16, and so on. In the 2nd rectangle whose area is 1/6, start by dividing in a ratio of 2 to 1 to make 1/9+1/18. Then divide the smaller part again in a ratio of 2 to 1 to get 1/9+1/27+1/54; and so on. In the 3rd rectangle divide in a ratio of 3:4 each time to get 1/12 = 1/16+1/48 = 1/16+1/64+1/192 = ... And so on for all the rest of the rectangles. This isn't what I would call "trivial", because it only makes the "sum {n>=2} 1/(n-1) - 1/n = 1" part obvious, and does nothing for the "sum {k>=2} 1/n^k = 1/n(n-1)" part. (There is another drawing for that, but I couldn't see a nice way to work it into this drawing). But it's visual and would probably make a nice colour illustration for a webpage (-: - Robert On Tue, Jan 3, 2012 at 22:00, Gareth McCaughan <gareth.mccaughan@pobox.com>wrote:

[...]

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?)

-- 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