# ON A RELATION BETWEEN THE RIEMANN ZETA FUNCTION AND Hirotaka Sato ```INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008), #A53
ON A RELATION BETWEEN THE RIEMANN ZETA FUNCTION AND
THE STIRLING NUMBERS
Hirotaka Sato
Japanese Language Center for International Students, Tokyo University of Foreign Studies, 3-11-1,
Asahi-cho, Fuchu, Tokyo, 183-8534, Japan
htsato@tufs.ac.jp
Received: 12/17/07, Accepted: 11/3/08, Published: 12/3/08
Abstract
Let ζ(z) be the Riemann zeta function and s(k, n) the Stirling numbers of the first kind.
!
s(k,n)
Shen proved the identity ζ(n + 1) = ∞
(1 ≤ n ∈ Z). We give a short proof by
k=n k&middot;k!
elementary methods.
1. The Result
!∞ −z
Let ζ(z) =
be the Riemann zeta function, and let s(k, n) denote the Stirling
k=1 k
numbers of the first kind, which are defined by
s(0, 0) = 1 , s(k, 0) = s(0, n) = 0
s(k + 1, n + 1) = s(k, n) + k &middot; s(k, n + 1)
(k =
# 0, n #= 0),
(k ∈ Z, n ∈ Z).
(1)
(2)
Shen  proved the following identity, which shows an interesting relation between ζ(n) and
s(k, n) by using Gauss’s summation theorem of the hypergeometric series:
ζ(n + 1) =
∞
&quot;
s(k, n)
k=n
k &middot; k!
(1 ≤ n ∈ Z).
(3)
In this paper we give a short proof of (3) by elementary methods.
First we show the outline of the proof. We denote
1
(k)−n =
(1 ≤ n ∈ Z, 1 ≤ k ∈ Z)
k(k + 1)(k + 2) &middot; &middot; &middot; (k + n − 1)
!
and put ξ(n) = ∞
k=1 (k)−n . Then we have
ξ(n + 1) =
∞
&quot;
1
1
{(k)−n − (k + 1)−n } =
.
n
n
&middot;
n!
k=1
(4)
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008), #A53
Proposition. For 1 ≤ x ∈ R and 0 ≤ n ∈ Z we have
∞
&quot;
−(n+1)
x
=
s(k, n) &middot; (x)−(k+1) .
2
(5)
k=n
By this proposition we have
ζ(n + 1) =
∞
&quot;
−(n+1)
m
=
m=1
∞ &quot;
∞
&quot;
m=1 k=n
s(k, n) &middot; (m)−(k+1) .
Since it is a convergent series with positive terms, we can change the order of summation.
Noting (4), we obtain
∞
∞
&quot;
&quot;
s(k, n)
ζ(n + 1) =
s(k, n)ξ(k + 1) =
.
k
&middot;
k!
k=n
k=n
Now we prove the proposition above. We need the following result [1, Section 54, p. 160].
Lemma.
s(N, k)
.
N →∞
N!
For fixed 1 ≤ k ∈ Z, we have lim
!
We prove (5) by induction on n. The case n = 0, which is x−1 = ∞
k=0 s(k, 0) &middot; (x)−(k+1) ,
follows from (1) and the definition of (x)−k . Now let N be a sufficiently large integer. From
(2) we have
N
N
&quot;
&quot;
s(k, n) &middot; (x)−(k+1) =
(s(k + 1, n + 1) − k &middot; s(k, n + 1)) &middot; (x)−(k+1)
k=n
k=n
=
=
=
N
&quot;
k=n
N
&quot;
s(k + 1, n + 1) &middot; (x)−(k+1) −
k=n
k &middot; s(k, n + 1) &middot; (x)−(k+1)
s(k + 1, n + 1) &middot; (x)−(k+2) &middot; (x + k + 1) −
k=n
N
+1
&quot;
k=n+1
s(k, n + 1) &middot; (x)−(k+1) &middot; (x + k) −
(Note s(n, n + 1) = 0.)
=x&middot;
N
&quot;
N
+1
&quot;
k=n+1
N
&quot;
k=n+1
N
&quot;
k=n
k &middot; s(k, n + 1) &middot; (x)−(k+1)
k &middot; s(k, n + 1) &middot; (x)−(k+1)
s(k, n + 1) &middot; (x)−(k+1) + s(N + 1, n + 1) &middot; (x)−(N +2) &middot; (N + 1).
Noting x ≥ 1, we obtain
N +1
1 &middot; 2 &middot; &middot; &middot; (N + 2)
s(N + 1, n + 1) N + 1
=
&middot;
,
(N + 1)!
N +2
s(N + 1, n + 1) &middot; (x)−(N +2) &middot; (N + 1) ≤ s(N + 1, n + 1) &middot;
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008), #A53
3
which tends to 0 as N → ∞ because of the lemma above. Therefore as N → ∞ we obtain
by the induction assumption
−(n+1)
x
=x&middot;
∞
&quot;
k=n+1
s(k, n + 1) &middot; (x)−(k+1) .
Hence we have complete the proof of (5).
Remark. Let S(n, k) be the Stirling number of the second kind and denote (x)n = x(x −
1) &middot; &middot; &middot; (x − n + 1) for 1 ≤ n ∈ Z. Equation (5) can be viewed as the negative n case of the
well-known identity
n
&quot;
n
x =
S(n, k) &middot; (x)k (0 ≤ n ∈ Z).
k=0
References
 C. Jordan, Calculus of Finite Differences, 3rd ed., Chelsea, 1965.
 L. C. Shen, Remarks on some integrals and series involving the Stirling numbers and ζ(n), Trans. Amer.
Math. Soc. 347 (1995), 1391-1399.
```