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Question Number 160496 by mathlove last updated on 30/Nov/21
Answered by Kamel last updated on 01/Dec/21
z2n+1−1=∏2nk=0(z−e2kπi2n+1)∴∑2nk=0zk=∏2nk=1(z−e2kπi2n+1)So:forz=1,2n+1=∏2nk=1(e2kπi2n+1−1)Forz=−1,1=∏2nk=1(e2kπi2n+1+1)tan(kπ2n+1)=e2kπi2n+1−1i(e2kπi2n+1+1)∴∏2nk=1(−1)ntan(kπ2n+1)=2n+1=p=2n+1−k(−1)n∏nk=1tan(kπ2n+1)∏np=1tan(π−pπ2n+1)Then:2n+1=(∏nk=1tan(kπ2n+1))2Or:tan(π2n+1)tan(2π2n+1)...tan(nπ2n+1)=2n+1
Answered by mr W last updated on 30/Nov/21
(2)limn→∞∑nk=11n+k=limn→∞1n∑nk=11kn+1=∫01dxx+1=ln(x+1)∣01=ln2
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