Value of zeta(2)

Here shows that

using the sine factorization formula. The session file can be obtained from here.

Maxima 5.20.0 http://maxima.sourceforge.net
using Lisp CMU Common Lisp 19f (19F)
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) block(load(("/Users/yasube/Programming/test-imaxima/cmucl/5.20.0/share/maxima/5.20.0/emacs/imaxima.lisp")), linenum:0)$

(%i1) powerdisp:true;


(%i2) sin(%pi*x)=%pi*x*product(1-x^2/n^2,n,1,inf);


(%i3) sin(%pi*x)=taylor(sin(%pi*x),x,0,10);


(%i4) coeff(rhs(%),x,3);

We see how the first n terms of the infinite product is expanded with n=2,3,4,5,6, and 10 and seet the coincidence of coeffcients of x^3 to the sum of first n terms of the infinite sum definition of zeta function.

(%i5) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=2;


(%i6) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=2;


(%i7) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=3;


(%i8) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=3;


(%i9) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=4;


(%i10) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=4;


(%i11) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=5;


(%i12) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=5;


(%i13) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=6;


(%i14) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=6;


(%i15) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=10;


(%i16) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=10;

Thus the following equation holds:

(%i17) -sum(1/n^2,n,1,inf)*%pi=-%pi^3/6;


(%i18) %/(-%pi);


(%i19)