You're trying to get a gamma function and pi/6 to come out of it? I see you took it through a couple binomial expansions (...right?), can you use Stirling's Approximation to bridge the gap?
Ah, sorry. That's not gamma function but Euler–Mascheroni constant that I forgot to specify. And only now I can see how in the last row of my calculations I'm getting explicit formula for γ²… so I think that I can call the problem as solved :D. Thanks! Regarding Stirling's Approximation: I don't know if I can use it here, but it feels like a method that I should read-up on.
Glad to hear your problem was already solved, heh. Yeah I was hesitant to bring up Stirling's Approximation, you probably can't use approximations during strict proofs. Keep math-ing, good sir Devac, and thanks for introducing me to mathb.in, looks like a great website.
Well, it got solved mainly because I had to look-up the definition ;). I was and am by all means sincere with my thanks. Regarding approximations in strict proofs, it's sometimes a must. Mathematical physics example would be derivation of Einstein's formula for E = mc², which in this case is as strict as you can get while being simply aware that higher orders don't carry much relevance. Purely mathematical example would be pretty much any proof that requires showing how something looks close to boundary or critical point. Specific example: 2 (sin(x) + sin(3x)/3 + sin(5x)/5 + … ) = 2 \sum_{k = 1}^{\infty}{\frac{sin(2k - 1)x}{2k - 1}} Sum for 0 < x < π is π/2 Sum for x = 0 or x = π or x = -π is 0 Sum for -π < x < 0 is -π/2 So we can now see that sums experience two 'jumps' (or discontinuities): S(+0) - S(0) = π/2 S(0) - S(-0) = π/2 To investigate, we have to carefully check partial sums. Basically whole procedure is all about applying approximation and step-by-step getting to analytical results. Same works for any general function that obeys above criteria, only does not offer as quick and easy way to see it (personal opinion).