Problem 11456
(American Mathematical Monthly, Vol.116, October 2009) Proposed by R. Mortini (France).
Find
n→∞lim n
n
Y
m=1
1 − 1
m+ 5 4m2
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Let z = −21+ i then
eγz
n
Y
m=1
1 + z m
e−z/m·eγz
n
Y
m=1
1 + z
m
e−z/m= eHn−γ
n
Y
m=1
1 +
z m
2= eHn−γ
n n
n
Y
m=1
1 − 1
m+ 5 4m2
where Hn =Pn k=11
k. Since Hn= log n + γ + o(1) then eHn−γ/n → 1.
Moreover, by Gamma function’s definition 1
zΓ(z) = eγz
∞
Y
m=1
1 + z m
e−z/m.
Hence
n→∞lim n
n
Y
m=1
1 − 1
m+ 5 4m2
= 1
zΓ(z)· 1
zΓ(z) = 1
Γ(z + 1)· 1 Γ(z + 1). Finally, since z + 1 = 12+ i and z + 1 = 12 −i = 1 − (z + 1), by the reflection property
n→∞lim n
n
Y
m=1
1 − 1
m + 5 4m2
=sin(π(z + 1))
π = cos(iπ)
π =cosh(π) π .