Hi dear James, yes, looking at it, there may be confusion. The approach below with the limits is much clearer. Pi=Limit(2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((2-1/(cos(Pi*2^(-i-2)))^2)^(2^(i-1)),i,1,n))^(1/2^n))),n=infinity); The same be true for the other products in the list. best regards. Le Mardi 31 octobre 2017 19h00, "math-fun-request@mailman.xmission.com" <math-fun-request@mailman.xmission.com> a écrit : Send math-fun mailing list submissions to     math-fun@mailman.xmission.com To subscribe or unsubscribe via the World Wide Web, visit     https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun or, via email, send a message with subject or body 'help' to     math-fun-request@mailman.xmission.com You can reach the person managing the list at     math-fun-owner@mailman.xmission.com When replying, please edit your Subject line so it is more specific than "Re: Contents of math-fun digest..." Today's Topics:   1. Some formulas of pi in infinite product:       (Fran?ois Mendzina Essomba) ---------------------------------------------------------------------- Message: 1 Date: Tue, 31 Oct 2017 11:26:58 +0100 From: Fran?ois Mendzina Essomba <m_essob@yahoo.fr> To: math-fun@mailman.xmission.com, m_essob@yahoo.fr Subject: [math-fun] Some formulas of pi in infinite product: Message-ID: <5044a46f-24cf-1b86-7252-5d15c807f852@yahoo.fr> Content-Type: text/plain; charset=utf-8; format=flowed Hello, Some formulas of pi in infinite product: (4/Pi)=product(1/(1-tan(Pi*2^(-n-3))^2),n=0..infinity); (4/Pi)=product((1/2)*(1+1/cos(Pi*2^(-n-2))),n=0..infinity); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((2-1/(cos(Pi*2^(-n-2)))^2)^(2^(n-1)),n=1..infinity))^(1/2^n))); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((1-tan(Pi*2^(-n-2))^2)^(2^(n-1)),n=1..infinity))^(1/2^n))); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product(2^(2^(n-1))/(1/cos(Pi*2^(-n-1))+1)^(2^(n-1)),n=1..infinity))^(1/2^n))); Pi/4=sqrt(3*2^(2*n+1)-sqrt(3)*2^(4*n+2)*sqrt(2^(-2^(-n-1)-4*n-1)/(product((1-tan(Pi*2^(-n-2))^2)^(2^(n-1)),n=1..infinity))^(1/2^n)+2^(-4*n-2))); Pi/4=sqrt(3*2^(2*n+1)-sqrt(3)*2^(4*n+2)*sqrt(2^(-2^(-n-1)-4*n-1)/(product((2-1/(cos(Pi*2^(-n-2)))^2)^(2^(n-1)),n=1..infinity))^(1/2^n)+2^(-4*n-2))); 4/Pi=product(((cos((2*Pi)/8^n)+cos(Pi/8^n))*sec((2*Pi)/8^n))/8+(cos((3*Pi)/(2*8^n))*sec((2*Pi)/8^n))/4+((cos(Pi/8^n)+1)*sec((2*Pi)/8^n))/8+(cos(Pi/(2*8^n))*sec((2*Pi)/8^n))/4,n=1..infinity); Best regards. ------------------------------ Subject: Digest Footer _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ------------------------------ End of math-fun Digest, Vol 176, Issue 41 *****************************************