Question Number 9458 by geovane10math last updated on 09/Dec/16
$$\mathrm{If}\:\mathrm{the}\:\mathrm{zeta}\:\mathrm{function}\:\mathrm{of}\:\mathrm{2}\:\mathrm{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\zeta}\left(\mathrm{2}\right)\:=\:\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\boldsymbol{{n}}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\zeta}\left(\mathrm{2}\right)\:=\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{infinite}\:\boldsymbol{\mathrm{rational}}\:\mathrm{numbers}, \\ $$$$\mathrm{why}\:\mathrm{converges}\:\mathrm{for}\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}},\:\mathrm{an}\:\boldsymbol{\mathrm{irrational}} \\ $$$$\mathrm{number}? \\ $$
Commented by geovane10math last updated on 09/Dec/16
$${Know}\:{if}\:{an}\:{infinite}\:{serie}\:{diverges}\:{or} \\ $$$${converges}\:{is}\:{easy}... \\ $$$${How}\:{calculate}\:{the}\:{value}\:{of}\:{serie}? \\ $$