Wouter> rwg>And, as Neil empiricizes, Sum[2^k/(1 + z^2^k), {k, 0, Infinity}] == 1/(z-1) 1. Why did Julian change his name? 2. do I understand your "it telescopes" correctly as f[z_,n_]:=Sum[2^k/(1 + z^2^k), {k, 0, n}] implies f[z, n+1] == 1/(1 + z) + 2*f[z^2, n] (1) 3. it is easy to show that 1/(z-1) satisfies (1) for n->inf and thus dropped, but how do you generaly solve a recursion like that? Look at it until the solution occurs to you? Wouter. 1. I have three tutors: Neil Bickford, http://nbickford.wordpress.com/ (who is currently trying to 3Dprint a largish biroller, and should soon announce his sliding block results) Julian Ziegler Hunts, the "evil child" who cooked DanA's extension of Alon Amit's vector puzzle, coauthor of http://www.blurb.com/b/2172660-minskys-trinskys-3rd-edition Corey Ziegler Hunts, author of the Mma version of my path invariant matrix tools and a really handy radical denester, and coauthor of Minskys & Trinskys. 2. It's simpler than that: FullSimplify[ 2^(1 + n)/(1 - x^2^(1 + n)) -(2^n/(1 - x^2^n))]== 2^n/(1 + x^2^n), so f[z,n]==1/(1-z) - 2^(n+1)/(1-z^2^(n+1)), giving both infinite sums. And the above identity follows from (f[z,oo]==1/(z-1)) - (f[z^2^(n+1),oo] == 1/(z^2^(n+1)-1), so it practically proves itself. 3. If we knew how to do this, we would teach our computer algebra systems. WHOA! Mma (9.0.1) can do this! In[80]:= RSolve[f[n + 1] - f[n] == 2^n/(1 + x^2^n), f[n], n] Out[80]= {{f[n] -> 1/(-1 + x) - 2^n/(-1 + x^2^n) + C[1]}} Now all they need to do is call RSolve from Sum. And all we need to do is figure out how they did it. DanA>Both of these are incredibly beautiful identities, especially if true. Which, given their source, seems very likely. If anyone has proved this already, my guess is it would be Euler. Have his complete works been collected yet, and even better, are they freely available in English translation on the Web? --Dan On 2013-02-27, at 3:05 PM, Bill Gosper wrote: Sum[2^k/(1 + z^2^k), {k, -Infinity, Infinity}] == 1/Log[z] Can somebody tell me where? --rwg And, as Neil empiricizes, Sum[2^k/(1 + z^2^k), {k, 0, Infinity}] == 1/(z-1) I clearly remember confronting several shelf-feet of Euler's Opera Omnia in the Stanford Math Library, while working for Knuth. The Latin is only half the problem, since Euler was only part way through inventing all our familiar notation! That guy was inhuman. He even knew about the motion of a freely tumbling brick. So I'd be surprised if he *didn't* know these z^2^n identities. But it would be nice to know for sure their origin. --rwg