>>304
マクローリン展開
　Σ[k=1,∞] (1/k)x^(k-1) = -(1/x)log(1-x),
より
　Σ[k=1,∞] 1/(kk･2^k) = -∫[0〜1/2] (1/x)log(1-x) dx,
　Σ[k=1,∞] {1/kk - 1/(kk･2^k)} = -∫[1/2〜1] (1/y)log(1-y) dy,
辺々引く。
　ζ(2) - Σ[k=1,∞] 2/(kk･2^k)
　= -∫[1/2〜1] log(1-y)/y dy + ∫[0〜1/2] (1/x)log(x) dx,
　= -∫[0〜1/2] log(x)/(1-x) dx + ∫[0〜1/2] (1/x)log(1-x) dx
　= [ log(x)log(1-x) ](x=0,1/2)
　= (log 1/2)^2
　= (log 2)^2
　= 0.4804530139182

http://club.informatix.co.jp/?p=3326

数列総合スレ
http://rio2016.5ch.net/test/read.cgi/math/1290234907/203-205

オイラーの贈物スレ
http://rio2016.5ch.net/test/read.cgi/math/1417406099/244ｰ247