I'm pretty sure Gareth nailed the sum I was after. And I agree with him that the striking fact that (*) Sum_{k,n=2..oo} 1/k^n = 1 cries out for an immediate explanation, even though Gareth's computational proof can hardly be shortened. Meanwhile, I'm also sure I recall recently reading of something related being = 3/5. How about this: What is Sum_{1/L | L = k^n for some k,n >= 2} ? In other words, the same sum as (*) but with all repetitions removed. --Dan << Gareth wrote: << I 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?)

...

________________________________________________________________________________________ It goes without saying that .