Infinite Products Involving ζ(3) and Catalan’s Constant

Yasuyuki Kachi Department of Mathematics University of Kansas Lawrence, KS 66045-7523 USA

                   kachi@math.ku.edu

Pavlos Tzermias Department of Mathematics University of Patras 26500 Rion (Patras) Greece

               tzermias@math.upatras.gr

Abstract

We present some infinite product formulas for e^{\frac{7ζ(3)}{π^{2}}} , e^{\frac{4G}{π}} and e^{\frac{2G}{π} ± \frac{1}{2}} , where G is Catalan’s constant. We relate these formulas to similar ones obtained by Guillera and Sondow in the context of their systematic study of Lerch’s transcendent. Our proofs are entirely elementary.

1 Introduction

This paper studies some infinite product formulas involving two classical constants, namely ζ(3) and Catalan’s constant, whose definition we now recall:

ζ(3) = ∑_{n=1}^{∞} \frac{1}{n^{3}}

and

G = ∑_{m=0}^{∞} \frac{( − 1)^{m}}{(2m + 1)^{2}} .

The following formulas are reminiscent of similar formulas obtained by Guillera and Sondow in [5]:

Proposition 1. The following formulas hold:

e^{\frac{7ζ(3)}{4π^{2}} + \frac{1}{4}} = lim_{m→∞} ∏_{n=1}^{2m+1} \frac{1}{\sqrt[4]{e}} (1 − \frac{1}{n + 1})^{\frac{n(n+1)}{2} (−1)^{n}}.  (1)

e^{\frac{7ζ(3)}{4π^{2}} − \frac{1}{4}} = lim_{m→∞} ∏_{n=1}^{2m} \sqrt[4]{e} (1 − \frac{1}{n + 1})^{\frac{n(n+1)}{2} (−1)^{n}}.  (2)

e^{\frac{7ζ(3)}{π^{2}}} = lim_{m→∞} (\frac{2^{2^{2}} · 4^{4^{2}} · 6^{6^{2}} ⋯ (2m)^{(2m)^{2}}}{1^{1^{2}} · 3^{3^{2}} · 5^{5^{2}} ⋯ (2m − 1)^{(2m−1)^{2}}} )^{4} (\frac{(2m + 2)^{4m+5}}{(2m + 1)^{12m+9}} )^{m}.  (3)

Proposition 2. The following formulas hold:

e^{\frac{2G}{π} − \frac{1}{2}} = lim_{m→∞} ∏_{n=1}^{2m} (1 − \frac{2}{2n + 1})^{n(−1)^{n}}.  (4)

e^{\frac{2G}{π} + \frac{1}{2}} = lim_{m→∞} ∏_{n=1}^{2m+1} (1 − \frac{2}{2n + 1})^{n(−1)^{n}}.  (5)

e^{\frac{4G}{π}} = lim_{m→∞} (\frac{3^{3} · 7^{7} · 11^{11} ⋯ (4m − 1)^{4m−1}}{1^{1} · 5^{5} · 9^{9} ⋯ (4m − 3)^{4m−3}} )^{2} \frac{(4m + 3)^{2m+1}}{(4m + 1)^{6m+1}} .  (6)

We claim no novelty for the formulas themselves; our only purpose here is to present completely elementary proofs of these formulas and to establish the not-so-obvious facts below:

Fact 3. Formula (3) is equivalent to the following formula given by Guillera and Sondow [5, Example 5.3]:

e^{\frac{7ζ(3)}{4π^{2}}} = e ^{∑_{n=1}^{∞} \frac{n(n+1)}{2^{n+3}} ∑_{k=0}^{n} (−1)^{k+1} \binom{n}{k} log(k+1)}

= ∏_{n=1}^{∞} ( ∏_{k=0}^{n} (k + 1)^{(−1)^{k+1} \binom{n}{k}})^{\frac{n(n+1)}{2^{n+3}}}

= (\frac{2^{1}}{1^{1}} )^{\frac{1·2}{2^{4}}} (\frac{2^{2}}{1^{1} · 3^{1}} )^{\frac{2·3}{2^{5}}} (\frac{2^{3} · 4^{1}}{1^{1} · 3^{3}} )^{\frac{3·4}{2^{6}}} (\frac{2^{4} · 4^{4}}{1^{1} · 3^{6} · 5^{1}} )^{\frac{4·5}{2^{7}}} · · · .

Fact 4. Formula (4) follows from rearranging the factors of the following formula given by Guillera and Sondow [5, Example 5.5]:

e^{\frac{G}{π}} = e ^{∑_{n=1}^{∞} \frac{n}{2^{n+2}} ∑_{k=0}^{n} (−1)^{k+1} \binom{n}{k} log(2k+1)}

= ∏_{n=1}^{∞} ( ∏_{k=0}^{n} (2k + 1)^{(−1)^{k+1} \binom{n}{k}})^{\frac{n}{2^{n+2}}}

= (\frac{3^{1}}{1^{1}} )^{\frac{1}{2^{3}}} (\frac{3^{2}}{1^{1} · 5^{1}} )^{\frac{2}{2^{4}}} (\frac{3^{3} · 7^{1}}{1^{1} · 5^{3}} )^{\frac{3}{2^{5}}} (\frac{3^{4} · 7^{4}}{1^{1} · 5^{6} · 9^{1}} )^{\frac{4}{2^{6}}} ⋯,

which in turn is equivalent to formula (6).

2 Proof of Proposition 2

We begin with the following formula which is a classically known Fourier expansion (see, for example, Exercise 11.15(c) in [1, p. 338]):

Formula 5. Let σ ∈ [ − \frac{1}{2}, \frac{1}{2}] \ \{0\}. Then

∑_{m=0}^{∞} \frac{cos (π(2m + 1) σ)}{2m + 1} = \frac{1}{2} log | cot (\frac{π}{2} σ) |.

The following formula, which follows directly from Formula 5 by integrating both sides over the interval [0, \frac{1}{2}], is also well-known (see, for example, [2, p. 239]):

Formula 6.

G = ∫_{θ=0}^{π/4} log ( cot θ) dθ.

By applying integration by parts to the latter integral, we obtain

Corollary 7.

G = \frac{1}{2} ∫_{α=0}^{1/2} \frac{π^{2} α}{sin (π α)} dα.

The following formula is also well-known (see, for example, [8, p. 155]):

Formula 8. Let α ∈ ℝ \ ℤ and s ∈ ( − \frac{1}{2}, \frac{1}{2}). Then

cos (2παs) = \frac{sin (πα)}{π} (\frac{1}{α} + 2α ∑_{m=1}^{∞} \frac{( − 1)^{m}}{α^{2} − m^{2}} cos (2πms)).

Setting s = 0 in Formula 8 gives:

Corollary 9. Let α ∈ ℝ \ ℤ. Then

1 = \frac{sin (πα)}{π} (\frac{1}{α} + 2α ∑_{m=1}^{∞} \frac{( − 1)^{m}}{α^{2} − m^{2}} ).

Lemma 10. Let m ∈ ℤ, m ≥ 1. Then

∫_{α=0}^{1/2} \frac{α^{2}}{α^{2} − m^{2}} dα = \frac{1}{2} + \frac{m}{2} log \frac{2m − 1}{2m + 1}.

Proof. This is straightforward:

∫_{α=0}^{1/2} \frac{α^{2}}{α^{2} − m^{2}} dα = \frac{1}{2} ∫_{α=0}^{1/2} (2 + \frac{m}{α − m} − \frac{m}{α + m}) dα

= \frac{1}{2} [2α + m log ( − α + m) − m log (α + m) ]_{α=0}^{1/2}

= \frac{1}{2} + \frac{m}{2} log \frac{2m − 1}{2m + 1}.

We now proceed with the proof of formula (4). By Corollary 9, we have

\frac{π^{2}α}{sin (πα)} = π + 2πα^{2} ∑_{m=1}^{∞} \frac{( − 1)^{m}}{α^{2} − m^{2}} .

Integrating both sides with respect to α over the interval [0, \frac{1}{2}] gives

∫_{α=0}^{1/2} \frac{π^{2}α}{sin (πα)} dα = \frac{π}{2} + 2π ∫_{α=0}^{1/2} ( ∑_{m=1}^{∞} \frac{(−1)^{m} α^{2}}{α^{2} − m^{2}} ) dα.  (7)

Consider the sequence of functions

f_{m}(α) = \frac{( − 1)^{m} α^{2}}{α^{2} − m^{2}}

on the interval I = [0, \frac{1}{2}], where m = 1, 2, …. Since α ∈ I, we clearly have

|f_{m}(α)| = \frac{α^{2}}{|α^{2} − m^{2}|} ≤ \frac{\frac{1}{4}}{m^{2} − \frac{1}{4}} = \frac{1}{4m^{2} − 1} ≤ \frac{1}{2m^{2}} ,

for all m. Since

∑_{m=1}^{∞} \frac{1}{2m^{2}}

converges, it follows from the Weierstrass M-test that the series

∑_{m=1}^{∞} f_{m}(α)

converges uniformly on I, and, by well-known principles, (see, for example, [1, Thm. 9.9, p. 226]), can therefore be integrated term by term. In other words, if we set

a_{m} = ( − 1)^{m} (1 + m log \frac{2m − 1}{2m + 1}),

then (7) and Lemma 10 imply that

∫_{α=0}^{1/2} \frac{π^{2}α}{sin (πα)} dα = \frac{π}{2} + π ∑_{m=1}^{∞} a_{m}.  (8)

The left-hand side of (8) is a definite integral of the continuous function \frac{π^{2}α}{sin(πα)} over the

interval [0, \frac{1}{2}]. Hence the left-hand side of (8) is a real number which implies that

lim_{m→∞} a_{m} = 0.

Keeping this in mind, define

A_{n} = ∑_{m=1}^{n} a_{m}.

Then, by (8), we have

∫_{α=0}^{1/2} \frac{π^{2}α}{sin (πα)} dα = \frac{π}{2} + π lim_{n→∞} A_{n} = \frac{π}{2} + π lim_{N→∞} A_{2N} = \frac{π}{2} + π lim_{N→∞} ∑_{m=1}^{N}(a_{2m−1} + a_{2m})

= π (\frac{1}{2} + lim_{N→∞} ∑_{m=1}^{N} ( − (2m − 1) log \frac{4m − 3}{4m − 1} + 2m log \frac{4m − 1}{4m + 1}))

= π (\frac{1}{2} + lim_{N→∞} log ∏_{m=1}^{N} \frac{(4m − 1)^{4m−1}}{(4m − 3)^{2m−1} (4m + 1)^{2m}} )

= π (\frac{1}{2} + log ∏_{m=1}^{∞} \frac{(4m − 1)^{4m−1}}{(4m − 3)^{2m−1} (4m + 1)^{2m}} ).

By Corollary 7, the left-hand side equals 2G, therefore

\frac{2G}{π} − \frac{1}{2} = log ∏_{m=1}^{∞} \frac{(4m − 1)^{4m−1}}{(4m − 3)^{2m−1} (4m + 1)^{2m}} .

Therefore,

e^{\frac{2G}{π} − \frac{1}{2}} = lim_{m→∞} \frac{3^{3}}{1^{1} · 5^{2}} · \frac{7^{7}}{5^{3} · 9^{4}} · \frac{11^{11}}{9^{5} · 13^{6}} ⋯ \frac{(4m − 1)^{4m−1}}{(4m − 3)^{2m−1}(4m + 1)^{2m}}

= lim_{m→∞} \frac{3^{3} · 7^{7} · 11^{11} ⋯ (4m − 1)^{4m−1}}{5^{5} · 9^{9} · 13^{13} ⋯ (4m − 3)^{4m−3} · (4m + 1)^{2m}}

= lim_{m→∞} (\frac{1}{3})^{−1}(\frac{3}{5})^{2}(\frac{5}{7})^{−3} ⋯ (\frac{4m − 1}{4m + 1})^{2m} = lim_{m→∞} ∏_{n=1}^{2m} (1 − \frac{2}{2n + 1})^{n(−1)^{n}},

and this completes the proof of formula (4). Multiplying both sides of the latter formula by e and using the fact that

e = lim_{m→∞} (1 − \frac{2}{4m + 3})^{−(2m+1)}

gives formula (5). Finally, multiplying formulas (4) and (5) together and expanding gives formula (6).

3 Proof of Proposition 1

We will first prove formula (1).

Lemma 11. Let m ∈ ℕ and δ ∈ (0, \frac{1}{2}). Then

π^{2} ∫_{σ=δ}^{1/2} (\frac{1}{2} − σ) \frac{cos (π(2m + 1)σ)}{2m + 1} dσ

= \frac{cos (π(2m + 1)δ)}{(2m + 1)^{3}} + \frac{π(δ − \frac{1}{2}) sin (π(2m + 1)δ)}{(2m + 1)^{2}}.

Proof. This is straightforward integration by parts:

∫_{σ=δ}^{1/2} (\frac{1}{2} − σ) \frac{cos (π(2m + 1)σ)}{2m + 1} dσ

= [(\frac{1}{2} − σ)\frac{sin (π(2m + 1)σ)}{π(2m + 1)^{2}}]_{σ=δ}^{1/2} + ∫_{σ=δ}^{1/2} \frac{sin (π(2m + 1)σ)}{π(2m + 1)^{2}} dσ

= (δ − \frac{1}{2}) \frac{sin (π(2m + 1)δ)}{π(2m + 1)^{2}} − [\frac{cos (π(2m + 1)σ)}{π^{2}(2m + 1)^{3}}]_{σ=δ}^{1/2} ,

and the claim follows.

Corollary 12. Let m ∈ ℕ. Then

π^{2} ∫_{σ=0}^{1/2} (\frac{1}{2} − σ) \frac{cos (π(2m + 1)σ)}{2m + 1} dσ = \frac{1}{(2m + 1)^{3}} .

Proof. In Lemma 11, let δ → 0+.

We now recall the following basic formula:

∑_{m=0}^{∞} \frac{1}{(2m + 1)^{3}} = \frac{7}{8} ζ(3).  (9)

We will establish the following:

Formula 13.

ζ(3) = \frac{4}{7} π G − \frac{2}{7} π^{2} ∫_{σ=0}^{1/2} \frac{πσ^{2}}{sin (πσ)} dσ.

Proof. First, we may rewrite Formula 6 as

G = ∫_{σ=0}^{1/2} \frac{π}{2} log ( cot (\frac{π}{2} σ)) dσ.  (10)

Second, by (9) and Corollary 12, we have

\frac{7}{8} ζ(3) = ∑_{m=0}^{∞} π^{2} ∫_{σ=0}^{1/2} (\frac{1}{2} − σ) \frac{cos (π(2m + 1)σ)}{2m + 1} dσ.  (11)

Fix δ ∈ (0, \frac{1}{2}). For each n ∈ ℕ, define the function

F_{n}(σ) = ∑_{m=0}^{n} (\frac{1}{2} − σ) \frac{cos (π(2m + 1)σ)}{2m + 1}

on the interval I = [δ, \frac{1}{2}]. The sequence

{ ∑_{m=0}^{n} cos (π(2m + 1)σ)}_{n∈ℕ}

of functions is uniformly bounded on I by (2 sin(πδ))^{−1} (see [1, Formula (15), p. 198] or [6, Item 185.5, p. 316]), whereas the sequence

\{(\frac{1}{2} − σ) \frac{1}{2m + 1}\}_{m∈ℕ}

clearly tends monotonically to 0 uniformly on I. Hence by applying Dirichlet’s test for uniform convergence (see [1, Thm. 9.15, p. 230] or or [6, p. 347]), it follows that the sequence of functions F_{n}(σ) converges uniformly on I. Therefore, the series

∑_{m=0}^{∞} (\frac{1}{2} − σ) \frac{cos (π(2m + 1)σ)}{2m + 1}

can be integrated term by term on I. Hence, Lemma 11 establishes the following Formula 14.

π^{2} ∫_{σ=δ}^{1/2} ∑_{m=0}^{∞} (\frac{1}{2} − σ) \frac{cos (π(2m + 1)σ)}{2m + 1} dσ

= ∑_{m=0}^{∞} \frac{cos (π(2m + 1)δ)}{(2m + 1)^{3}} + π (δ − \frac{1}{2}) ∑_{m=0}^{∞} \frac{sin (π(2m + 1)δ)}{(2 m + 1)^{2}}.

Now take the limits of both sides of the latter formula as δ → 0+. By the Weierstrass M-test, both series on the right-hand side of Formula 14 are uniformly convergent series of functions of δ on the interval I = [δ, \frac{1}{2}]. Therefore, we can interchange limits and infinite sums on the right-hand side of Formula 14 (see [1, Thm. 9.7, p. 220]). By (11), it follows that

π^{2} ∫_{σ=0}^{1/2} ∑_{m=0}^{∞} (\frac{1}{2} − σ) \frac{cos (π(2m + 1)σ)}{2m + 1} = \frac{7}{8} ζ(3).  (12)

Combining (10), (12) and Formula 5 gives

\frac{7}{8} ζ(3) = π^{2} ∫_{σ=0}^{1/2} (\frac{1}{2} − σ) \frac{1}{2} log ( cot (\frac{π}{2} σ)) dσ

= \frac{π^{2}}{4} ( ∫_{σ=0}^{1/2} log ( cot (\frac{π}{2} σ)) dσ − ∫_{σ=0}^{1/2} 2σ log ( cot (\frac{π}{2} σ)) dσ)

= \frac{π}{2} G − \frac{π^{2}}{2} ∫_{σ=0}^{1/2} σ log ( cot (\frac{π}{2} σ)) dσ.

In short,

ζ(3) = \frac{4}{7} π G − \frac{4}{7} π^{2} ∫_{σ=0}^{1/2} σ log ( cot (\frac{π}{2} σ)) dσ.  (13)

Formula 13 now follows because

∫_{σ=0}^{1/2} σ log ( cot (\frac{π}{2} σ)) dσ

= [\frac{σ^{2}}{2} log ( cot (\frac{π}{2} σ))]_{σ=0}^{1/2} − ∫_{σ=0}^{1/2} \frac{σ^{2}}{2} \frac{1}{cot (\frac{π}{2} σ)} \frac{−1}{( sin (\frac{π}{2}σ))^{2}} \frac{π}{2} d σ

= 0 + ∫_{σ=0}^{1/2} \frac{σ^{2}}{2} \frac{1}{cos (\frac{π}{2}σ) sin (\frac{π}{2}σ)} \frac{π}{2} dσ = \frac{1}{2} ∫_{σ=0}^{1/2} \frac{πσ^{2}}{sin (πσ)} σ.

The following statement is similar to Lemma 10.

Lemma 15. Let m ∈ ℤ, m ≥ 1. Then

∫_{σ=0}^{1/2} \frac{σ^{3}}{σ^{2} − m^{2}} dσ = \frac{1}{8} + \frac{m^{2}}{2} log \frac{4m^{2} − 1}{4m^{2}}.

Proof. This is straightforward:

∫_{σ=0}^{1/2} \frac{σ^{3}}{σ^{2} − m^{2}} dσ = ∫_{σ=0}^{1/2} (σ + m^{2} \frac{σ}{σ^{2} − m^{2}} ) dσ

= [ \frac{1}{2}σ^{2} + m^{2} \frac{1}{2} log ( − σ^{2} + m^{2})]_{σ=0}^{1/2}

= \frac{1}{8} + \frac{m^{2}}{2} log \frac{m^{2} − \frac{1}{4}}{m^{2}} .

Lemma 16.

lim_{n→∞} \frac{e^{n}}{(1 + \frac{1}{n})^{n^{2}}} = lim_{n→∞} \frac{e^{−n}}{(1 − \frac{1}{n})^{n^{2}}} = \sqrt{e}.

Proof. By taking logarithms, it suffices to show that

lim_{n→∞} (n − n^{2} log (1 + \frac{1}{n})) = \frac{1}{2} = lim_{n→∞} ( − n − n^{2} log (1 − \frac{1}{n})).

This follows by substituting x = ±\frac{1}{n} in the Maclaurin series of the function log (1 + x) and using continuity.

We now proceed with the proof of formula (1). By Corollary 9, we have

\frac{π σ^{2}}{sin (πσ)} = σ + 2σ^{3} ∑_{m=1}^{∞} \frac{( − 1)^{m}}{σ^{2} − m^{2}} .

Integrating both sides with respect to σ over the interval [0, \frac{1}{2}] and using formula (4) (and its proof) and Lemma 15 gives

∫_{σ=0}^{1/2} \frac{πσ^{2}}{sin (πσ)} dσ = ∫_{σ=0}^{1/2} (σ + 2σ^{3} ∑_{m=1}^{∞} \frac{( − 1)^{m}}{σ^{2} − m^{2}} ) dσ

= ∫_{σ=0}^{1/2} σ dσ + 2 ∫_{σ=0}^{1/2} ( ∑_{m=1}^{∞} ( − 1)^{m} \frac{σ^{3}}{σ^{2} − m^{2}} ) dσ

= \frac{1}{8} + 2 ∑_{m=1}^{∞} ( − 1)^{m} ∫_{σ=0}^{1/2} \frac{σ^{3}}{σ^{2} − m^{2}} dσ

= \frac{1}{8} + 2 ∑_{m=1}^{∞} ( − 1)^{m} (\frac{1}{8} + \frac{m^{2}}{2} log \frac{4m^{2} − 1}{4m^{2}})

= \frac{1}{8} + ∑_{m=1}^{∞} ( − 1)^{m} (\frac{1}{4} + m^{2} log \frac{4m^{2} − 1}{4m^{2}}),

which equals

\frac{1}{8} + ∑_{ℓ=1}^{∞} ( − (\frac{1}{4} + (2ℓ − 1)^{2} log \frac{4(2ℓ − 1)^{2} − 1}{4(2ℓ − 1)^{2}})

+ (\frac{1}{4} + (2ℓ)^{2} log \frac{4(2ℓ)^{2} − 1}{4(2ℓ)^{2}}))

= \frac{1}{8} + ∑_{ℓ=1}^{∞} ( − (2ℓ − 1)^{2} log \frac{4(2ℓ − 1)^{2} − 1}{4(2ℓ − 1)^{2}}

+ (2ℓ)^{2} log \frac{4(2ℓ)^{2} − 1}{4(2ℓ)^{2}})

= \frac{1}{8} + ∑_{ℓ=1}^{∞} log (\frac{(4ℓ − 1)^{4ℓ−1} (4ℓ + 1)^{(2ℓ)^{2}} (4ℓ − 2)^{2(2ℓ−1)^{2}}}{(4ℓ)^{2(2ℓ)^{2}} (4ℓ − 3 )^{(2ℓ−1)^{2}}} )

= \frac{1}{8} + ∑_{ℓ=1}^{∞} log (\frac{(4ℓ − 1)^{4ℓ−1}}{(4ℓ − 3)^{2ℓ−1} (4ℓ + 1)^{2ℓ}} )

+ ∑_{ℓ=1}^{∞} log (\frac{(4ℓ + 1 )^{4ℓ^{2}+2ℓ} (4ℓ − 2)^{2(2ℓ−1)^{2}}}{(4ℓ)^{2(2ℓ)^{2}} (4ℓ − 3)^{4ℓ^{2}−6ℓ+2}} )

= \frac{2G}{π} − \frac{3}{8} + 2 ∑_{ℓ=1}^{∞} log (\frac{(4ℓ + 1)^{2ℓ^{2}+ℓ} (4ℓ − 2 )^{(2ℓ−1)^{2}}}{(4ℓ)^{(2ℓ)^{2}} (4ℓ − 3)^{2ℓ^{2}−3ℓ+1}} ).

Therefore, by Formula 13, it follows that

\frac{7}{4π^{2}} ζ(3) = \frac{3}{16} + log ∏_{ℓ=1}^{∞} \frac{(4ℓ)^{(2ℓ)^{2}} (4ℓ − 3)^{2ℓ^{2}−3ℓ+1}}{(4ℓ + 1)^{2ℓ^{2}+ℓ} (4ℓ − 2)^{(2ℓ−1)^{2}}} .  (14)

Now the latter infinite product can be written as

lim_{N→∞} ∏_{ℓ=1}^{N} 2^{4ℓ−1} \frac{(2ℓ)^{(2ℓ)^{2}}}{(2ℓ − 1)^{(2ℓ−1)^{2}}} \frac{(4(ℓ − 1) + 1)^{2(ℓ−1)^{2}+(ℓ−1)}}{(4ℓ + 1)^{2ℓ^{2}+ℓ}} ,

which equals

lim_{N→∞} (\frac{2}{4N + 1})^{2N^{2}+N} ∏_{ℓ=1}^{N} \frac{(2ℓ)^{(2ℓ)^{2}}}{(2ℓ − 1)^{(2ℓ−1)^{2}}}

= lim_{N→∞} \frac{2^{2N^{2}+N} (2N + 1)^{(2N+1)^{2}}}{(4N + 1)^{2N^{2}+N}} ∏_{ℓ=1}^{N} \frac{(2ℓ)^{(2ℓ)^{2}}}{(2ℓ + 1)^{(2ℓ+1)^{2}}}

= lim_{N→∞} (\frac{2^{2N^{2}+N} (2N + 1)^{(2N+1)^{2}}}{(4N + 1)^{2N^{2}+N}} \frac{2}{(2N + 2)^{(2N+1)(N+1)}}

× ∏_{ℓ=1}^{N} \frac{(2ℓ)^{(2ℓ+1)ℓ} (2ℓ + 2)^{(2ℓ+1)(ℓ+1)}}{(2ℓ + 1)^{(2ℓ+1)^{2}}} )

= lim_{N→∞} (e^{\frac{N+1}{2}} (\frac{4N + 2}{4N + 1})^{2N^{2}+N} (\frac{2N + 1}{2N + 2})^{(2N+1)(N+1)} \frac{2}{\sqrt{e}}

× ∏_{ℓ=1}^{N} \frac{(2ℓ)^{(2ℓ+1)ℓ} (2ℓ + 2)^{(2ℓ+1)(ℓ+1)}}{\sqrt{e} (2ℓ + 1)^{(2ℓ+1)^{2}}} ).

We claim that

lim_{N→∞} \frac{e^{\frac{N+1}{2}}}{(1 − \frac{1}{4N+2})^{(2N+1)N} (1 + \frac{1}{2N+1})^{(2N+1)(N+1)}} = e^{− \frac{3}{16}} .  (15)

Indeed, by Lemma 16, we have

lim_{N→∞} \frac{e^{N+ \frac{1}{4}}}{(1 + \frac{1}{2N+1})^{\frac{(2N+1)^{2}}{2}}} = 1 = lim_{N→∞} \frac{e^{\frac{1}{2}}}{(1 + \frac{1}{2N+1})^{\frac{2N+1}{2}}}

and

lim_{N→∞} \frac{e^{− \frac{N}{2} − \frac{5}{16}}}{(1 − \frac{1}{4N+2})^{\frac{(4N+2)^{2}}{8}}} = 1 = lim_{N→∞} \frac{e^{\frac{1}{4}}}{(1 − \frac{1}{4N+2})^{− \frac{4N+2}{4}}} ,

hence (15) follows. Combining (14) with (15) gives

e^{\frac{7ζ(3)}{4π^{2}}} = \frac{2}{\sqrt{e}} ∏_{l=1}^{∞} \frac{(2l)^{(2l+1)l} (2l + 2)^{(2l+2)(l+1)}}{\sqrt{e} (2l + 1)^{2l+1)^{2}}} .

Therefore,

e^{\frac{7ζ(3)}{4π^{2}}} = \frac{2}{\sqrt{e}} lim_{m→∞} ∏_{n=1}^{m} \frac{(2n)^{(2n+1)n} (2n + 2)^{(2n+2)(n+1)}}{\sqrt{e} (2n + 1)^{(2n+1)^{2}}}

= \frac{(2m + 2)^{(2m+1)(m+1)}}{e^{\frac{m+1}{2}}} \frac{2^{2^{2}} · 4^{4^{2}} · 6^{6^{2}} ⋯ (2m)^{(2m)^{2}}}{3^{3^{2}} · 5^{5^{2}} · 7^{7^{2}} ⋯ (2m + 1)^{(2m+1)^{2}}}

= e^{− \frac{1}{4}} lim_{m→∞} ∏_{n=1}^{2m+1} \frac{1}{\sqrt[4]{e}}(1 − \frac{1}{n + 1})^{\frac{n(n+1)}{2} (−1)^{n}}.

and this completes the proof of formula (1).

It remains to prove formulas (2) and (3). Note that

∏_{n=1}^{2m} \sqrt[4]{e} (1 − \frac{1}{n + 1})^{\frac{n(n+1)}{2} (−1)^{n}} = \frac{(1 − \frac{1}{2m+2})^{(2m+1)(m+1)}}{e^{−(m+ \frac{1}{4} )}} ∏_{n=1}^{2m+1} \frac{1}{\sqrt[4]{e}} (1 − \frac{1}{n + 1})^{\frac{n(n+1)}{2} (−1)^{n}}.

Therefore, formula (2) will follow from formula (1) once we show that

lim_{m→∞} \frac{e^{−(2m+ \frac{1}{2} )}}{(1 − \frac{1}{2m+2})^{(2m+2)(2m+1)}} = e.

This follows by writing (2m + 2)(2m + 1) as (2m + 2)^{2} − (2m + 2) and using Lemma 16. Now multiplying formulas (1) and (2) together and squaring gives

\begin{aligned}e^{\frac{7ζ(3)}{π^{2}}} & = lim_{m→∞} \frac{1}{\sqrt{e}} (\frac{2m + 1}{2m + 2})^{−(2m+1)(2m+2)} ∏_{n=1}^{2m} (1 − \frac{1}{n + 1})^{2n(n+1)(−1)^{n}} \\ & = lim_{m→∞} \frac{1}{\sqrt{e}} (\frac{(2m + 2)^{2m+2}}{(2m + 1)^{6m+2}} )^{2m+1} (\frac{2^{2^{2}} · 4^{4^{2}} · 6^{6^{2}} ⋯ (2m)^{(2m)^{2}}}{1^{1^{2}} · 3^{3^{2}} · 5^{5^{2}} ⋯ (2m − 1)^{(2m−1)^{2}}} )^{4} \end{aligned}

Formula (3) is now a consequence of the equality

\frac{1}{\sqrt{e}} = lim_{m→∞} (\frac{2m + 1}{2m + 2})^{m+2}.

4 Proof of Facts 3 and 4

By formula (3) and its proof, it suffices to show that the total exponent of k+1 in the infinite product expansion given by Guillera and Sondow [5, Example 5.3] equals ( − 1)^{k+1} (k + 1)^{2}, for all k ∈ ℕ. The exponent in question equals

( − 1)^{k+1} ∑_{n=k}^{∞} \binom{n}{k} \frac{n^{2} + n}{2^{n+3}} = \frac{( − 1)^{k+1}}{8 · (k!)} ∑_{n=k}^{∞} \frac{(n + 1) n^{2} (n − 1) ⋯ (n − k + 1)}{2^{n}}

= \frac{( − 1)^{k+1}}{8 · (k!)} ( ∑_{n=k}^{∞} \frac{(n + 2) (n + 1) ⋯ (n − k + 1)}{2^{n}} − 2 ∑_{n=k}^{∞} \frac{(n + 1) n ⋯ (n − k + 1)}{2^{n}})

= \frac{( − 1)^{k+1}}{8 · (k!)} (\frac{1}{2^{k}} ∑_{m=0}^{∞} \frac{(m + k + 2) (m + k + 1) ⋯ (m + 1)}{2^{m}}

− \frac{1}{2^{k−1}} ∑_{m=0}^{∞} \frac{(m + k + 1) (m + k) ⋯ (m + 1)}{2^{m}}).

We have the following lemma:

Lemma 17. For all k ∈ ℕ, we have

∑_{m=0}^{∞} \frac{(m + k + 1) (m + k) ⋯ (m + 1)}{2^{m}} = 2^{(k+2)} · ((k + 1)!).

Proof. This follows by term-by-term (k + 1)-fold differentiation of the geometric series

∑_{n=0}^{∞} x^{n} = (1 − x)^{−1}

and subsequent evaluation at x = \frac{1}{2}.

Therefore, by Lemma 17, the exponent in question equals

\frac{(−1)^{k+1}}{8 · (k!)} (8 · ((k + 2)!) − 8 · ((k + 1)!)) = (−1)^{k+1} (k + 1)^{2},

which completes the proof of Fact 3.

We will now show that, apart from the factor e^{− \frac{1}{2}} on the left-hand side of formula (4), the product expansion given by the latter formula and the product expansion given by Guillera and Sondow [5, Example 5.5] are equivalent. In other words, we will show that the total exponent of 2k + 1 in the infinite product expansion of e^{\frac{G}{π}} given by Guillera and Sondow [5, Example 5.5] equals (−1)^{k+1} (k + \frac{1}{2}), for all k ∈ ℕ. Since the infinite series involved is only conditionally convergent, the discrepancy involving e^{− \frac{1}{2}} can be explained by means of Riemann’s theorem on rearrangements of conditionally convergent series. The exponent in question equals

( − 1)^{k+1} ∑_{n=k}^{∞} \binom{n}{k} \frac{n}{2^{n+2}} = \frac{( − 1)^{k+1}}{4 · (k!)} ∑_{n=k}^{∞} \frac{n^{2} (n − 1) ⋯ (n − k + 1)}{2^{n}}

= \frac{( − 1)^{k+1}}{4 · (k!)} ( ∑_{n=k}^{∞} \frac{(n + 1) n ⋯ (n − k + 1)}{2^{n}} − ∑_{n=k}^{∞} \frac{n (n − 1) ⋯ (n − k + 1)}{2^{n}})

= \frac{( − 1)^{k+1}}{4 · (k!)} (\frac{1}{2^{k}} ∑_{m=0}^{∞} \frac{(m + k + 1) (m + k) ⋯ (m + 1)}{2^{m}}

− \frac{1}{2^{k}} ∑_{m=0}^{∞} \frac{(m + k) (m + k − 1) ⋯ (m + 1)}{2^{m}}).

By Lemma 17, this equals

\frac{(−1)^{k+1}}{4 · (k!)} (4 · ((k + 1)!) − 2 · (k!)) = ( − 1)^{k+1} (k + \frac{1}{2}) ,

as required, and this completes the proof of Fact 4.

5 Concluding remarks

Remark 18. The identities

∑_{n=k}^{∞} \binom{n}{k}\frac{n^{2} + n}{2^{n+3}} = (k + 1)^{2} ,  ∑_{n=k}^{∞} \binom{n}{k}\frac{n}{2^{n+2}} = k + \frac{1}{2}

which were used in the proofs of Facts 3 and 4 can also be very easily established by the Wilf-Zeilberger method via the use of Zeilberger’s Maple package EKHAD (see [9]).

Remark 19. One way to account for the fact that the products discussed in this paper are so closely tied to the ones studied by Guillera and Sondow in [5] is by noticing that they are related via Euler transformations. For instance, using the latter formula in the previous remark, one has

lim_{m→∞} ∑_{k=1}^{2m}(−1)^{k}k log \frac{2k − 1}{2k + 1} = lim_{m→∞} ∑_{k=1}^{2m}(−1)^{k} log \frac{2k − 1}{2k + 1} ∑_{n=k}^{∞} \binom{n}{k}\frac{n − 1}{2^{n+2}}.

If we interchange the summation on the right-hand side (an Euler transformation) the relation between formula (4) and the formula given in Fact 4 becomes evident.

Remark 20. The formulas in Propositions 2 and 1 are reminiscent of some powerful statements that deserve to be more widely known. We refer the reader to Finch’s book [4] for a wealth of information regarding such statements involving classical constants. For instance, the following function (first introduced by Borwein and Dykshoorn in [3]):

D(x) = lim_{m→∞} ∏_{n=1}^{2m+1} (1 + \frac{x}{n})^{n(−1)^{n+1}} = e^{x} lim_{m→∞} ∏_{n=1}^{2m} (1 + \frac{x}{n})^{n(−1)^{n+1}}.

Certain values of this function are related to some classical constants. Melzak proved in [7] that D(2) = \frac{πe}{2}. In [3], Borwein and Dykshoorn generalized Melzak’s result and explicitly determined the values of D(x) at all rational x having denominator 1, 2 or 3. Interestingly enough, some of the resulting evaluations involve Catalan’s constant, the Glaisher-Kinkelin constant and Γ(\frac{1}{4}). We have not been able to show that any of the formulas in Propositions 2 or 1 is a direct consequence of the latter evaluations.

6 Acknowledgment

We are much obliged to the referee for providing valuable references and insightful remarks on a previous version of this paper.

References

[1] T. Apostol, Mathematical Analysis, Addison Wesley, 1974.

[2] J. Borwein and D. Bailey, Mathematics by Experiment, A. K. Peters, 2004.

[3] P. Borwein and W. Dykshoorn, An interesting infinite product, J. Math. Anal. Appl., 179 (1993), 203–207.

[4] S. R. Finch, Mathematical Constants, Cambridge University Press, 2003.

[5] J. Guillera and J. Sondow, Double integrals and infinite products for some classical constants via analytic continuations of Lerch’s transcendent, Ramanujan J. 16 (2008), 243–270.

[6] K. Knopp, Theory and Application of Infinite Series, Dover, 1990.

[7] Z. A. Melzak, Infinite products for πe and π/e, Amer. Math. Monthly, 68 (1961), 39–41.

[8] P. Nahin, Dr. Euler’s Fabulous Formula: Cures Many Mathematical Ills, Princeton University Press, 2006.

[9] M. Petkŏvsek, H. Wilf, and D. Zeilberger, A = B, A. K. Peters, 1996.

2000 Mathematics Subject Classification: Primary 11Y60; Secondary 11M06, 33B99, 40A25. Keywords: Catalan’s constant, zeta(3), infinite product formula.

(Concerned with sequences \underline{A002117} and \underline{A006752}.)

Received October 1 2012; revised version received November 13 2012. Published in Journal of Integer Sequences, November 20 2012.

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