infinity series

We have the identity
(sin πz)/πz = Π(n=1 to ∞) (1-z2/n2)
(See Rudin, Real and Complex Analysis, Ch15, exercise 4)
Put z = i,
Π(n=1 to ∞) (1+1/n2) = (sin πi)/πi = (eπ-e-π)/2π.
So
Π(n=2 to ∞) (1+1/n2) = (eπ-e-π)/4π.


2009-03-18 22:00:05 補充：
這題太難了吧，不是專門學analysis，幾乎沒可能會懂。