1810 CHAPTER 57. INFINITE PRODUCTS
and therefore,∞
∑n=1
∣∣∣(1+zn
)e−zn −1
∣∣∣< ∞
with the convergence uniform on compact sets.
9. ↑ Show ∏∞n=1(1+ z
n
)e−zn converges to an analytic function on C which has zeros
only at the negative integers and that therefore,
∞
∏n=1
(1+
zn
)−1e
zn
is a meromorphic function (Analytic except for poles) having simple poles at thenegative integers.
10. ↑Show there exists γ such that if
Γ(z)≡ e−γz
z
∞
∏n=1
(1+
zn
)−1e
zn ,
then Γ(1) = 1. Thus Γ is a meromorphic function having simple poles at the negativeintegers. Hint: ∏
∞n=1 (1+n)e−1/n = c = eγ .
11. ↑ Now show that
γ = limn→∞
[n
∑k=1
1k− lnn
]
12. ↑Justify the following argument leading to Gauss’s formula
Γ(z) = limn→∞
(n
∏k=1
(k
k+ z
)e
zk
)e−γz
z
= limn→∞
(n!
(1+ z)(2+ z) · · ·(n+ z)ez(∑
nk=1
1k ))
e−γz
z
= limn→∞
n!(1+ z)(2+ z) · · ·(n+ z)
ez(∑nk=1
1k )e−z[∑n
k=11k−lnn]
= limn→∞
n!nz
(1+ z)(2+ z) · · ·(n+ z).
13. ↑ Verify from the Gauss formula above that Γ(z+1) = Γ(z)z and that for n a non-negative integer, Γ(n+1) = n!.
14. ↑ The usual definition of the gamma function for positive x is
Γ1 (x)≡∫
∞
0e−ttx−1dt.