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Question Number 152345 by mnjuly1970 last updated on 27/Aug/21
prove...S=∑∞n=11=13+2π327...◼
Answered by Kamel last updated on 27/Aug/21
prove...S=∑∞n=11=13+2π327...◼=∑+∞n=1nΓ(n+1)Γ(n)Γ(2n+1)=∑+∞n=1n∫01tn−1(1−t)ndt=∫01ddy[∑+∞n=1yn(1−t)n]y=tdt=∫011−t(1−t(1−t))2dt=∫01u(u2−u+1)2du=∫1+∞u(u2−u+1)2duS=12∫0+∞udu(u2−u+1)2=12dda[∫0+∞du(u2−au+1)]a=1=12dda[11−a24(π2+Arctan(a21−a24))]a=1=12[233(π2+Arctan(13))−123(−13−34)]=2π93+13=13+2π327
Commented by mnjuly1970 last updated on 28/Aug/21
verhnicesirkamel..tashakor
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