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Question Number 160496 by mathlove last updated on 30/Nov/21

Answered by Kamel last updated on 01/Dec/21

  z^(2n+1) −1=Π_(k=0) ^(2n) (z−e^((2kπi)/(2n+1)) )  ∴  Σ_(k=0) ^(2n) z^k =Π_(k=1) ^(2n) (z−e^((2kπi)/(2n+1)) )  So: for z=1, 2n+1=Π_(k=1) ^(2n) (e^((2kπi)/(2n+1)) −1)  For z=−1, 1=Π_(k=1) ^(2n) (e^((2kπi)/(2n+1)) +1)  tan(((kπ)/(2n+1)))=((e^((2kπi)/(2n+1)) −1)/(i(e^((2kπi)/(2n+1)) +1)))  ∴ Π_(k=1) ^(2n) (−1)^n tan(((kπ)/(2n+1)))=2n+1=^(p=2n+1−k) (−1)^n Π_(k=1) ^n tan(((kπ)/(2n+1)))Π_(p=1) ^n tan(π−((pπ)/(2n+1)))  Then: 2n+1=(Π_(k=1) ^n tan(((kπ)/(2n+1))))^2   Or: tan((π/(2n+1)))tan(((2π)/(2n+1)))...tan(((nπ)/(2n+1)))=(√(2n+1))

z2n+11=2nk=0(ze2kπi2n+1)2nk=0zk=2nk=1(ze2kπi2n+1)So:forz=1,2n+1=2nk=1(e2kπi2n+11)Forz=1,1=2nk=1(e2kπi2n+1+1)tan(kπ2n+1)=e2kπi2n+11i(e2kπi2n+1+1)2nk=1(1)ntan(kπ2n+1)=2n+1=p=2n+1k(1)nnk=1tan(kπ2n+1)np=1tan(πpπ2n+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)  lim_(n→∞) Σ_(k=1) ^n (1/(n+k))  =lim_(n→∞) (1/n)Σ_(k=1) ^n (1/((k/n)+1))  =∫_0 ^1 (dx/(x+1))  =ln (x+1)∣_0 ^1   =ln 2

(2)limnnk=11n+k=limn1nnk=11kn+1=01dxx+1=ln(x+1)01=ln2

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