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Question Number 195393 by Erico last updated on 01/Aug/23
provethatlimx→0∑nk=1(1−12k)xnx=14C2nnn
Answered by witcher3 last updated on 01/Aug/23
f(t)=tx=exln(t)=1+xln(t)+o(x),x→0∑nk=1(1−12k)x=Σ(1+xln(1−12k)+o(x))=n+xln(∏nk=1(2k−12k))+o(x)=n+xln(∏nk=1(2k)(2k−1)4k2)+o(x)=n+xln((2n)!4n(n!)2)+o(x)=n+xln(14nC2nn)+o(x)limx→01n∑nk=1(1−12k)xx=limex→01xln(1n(n+xln(14nC2nn)+o(x))=limex→0ln(1+xnln(14nC2nn)+o(1))x=elimx→0ln(1+xnln(C2nn4n)+o(1))x≪expiscontinus≫=elimx→0xnln(C2nn4n)+o(1)x=eln(C2nn4nn)=C2nn4nn14C2nnn
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