Geometric Moments and Rational Lambert Values: Proofs and Further Results

Will Cook

Long paper

Geometric Moments and Rational Lambert Values: Proofs and Further Results

Erdős #1049 By Will Cook September 2026 Authorship, AI use and citation

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Summary. The record proves the fixed-base Hankel asymptotic for Zudilin's normalised moments and the rational-base irrationality criterion log b/log a<0.4056830213840605…. The determinant proof separates geometric node configurations from positive weights; the irrationality proof instead specialises Zudilin's 2004 polynomial forms after cancellation. Coefficient moments, evaluated lattices and congruence conditions distinguish integer divisibility from nonzero remainder decay. A broader ratio-limit argument and a rational-value model remain unformalised. None of these arguments settles $F (3 / 2)$ .

This paper covers the sharp fixed-base Hankel asymptotic and rational-base irrationality proofs, coefficient moments and arithmetic obstructions, with ordinary ratio-limit and rational-value models distinguished from checked results.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #1049, which remains open.

Start here, by the editorial route

In this paper

The size of $V_{N}^{*}$ at a fixed base

The preceding bounds leave an error of order $N$ in the logarithm. We now determine its linear, logarithmic and constant terms. There are two calculations: compare every node configuration with the first $N$ nodes, then multiply the resulting moment weights. The first is independent of the particular weights under the hypotheses below; the second produces the Lambert-value correction for $V_{N}^{*}$ . Throughout this subsection $0 < q < 1$ is fixed, and we write

P = (q; q)_{\infty}, M (q) = \prod_{d \geq 1} (1 - q^{d})^{- d}, L = F (1 / q) = \sum_{r \geq 1} \frac{q^{r}}{1 - q^{r}},

with $B_{N} = \sum_{j < N} j^{2}$ and $C_{N} = \prod_{k < N} c_{k} = (N!)^{2} (N + 1)! / 2^{N}$ as above. Both products converge to positive finite values, and the abbreviation $L$ is used only in this subsection; it is the value at $q$ of the divisor series $L$ of Section 10. The estimates in this subsection concern real determinants at fixed $q$ . Denominator valuations for the 2004 forms were treated separately in Section 2.

For $a_{k} = 1$ , the moments are $(1 - q^{m + 1})^{- 1}$ , so Cauchy’s determinant formula gives the model calculation

\begin{matrix} (15) & D_{N}^{(0)} := det {(\frac{1}{1 - q^{i + j + 1}})}_{0 \leq i, j < N} = \frac{q^{B_{N}} Δ_{N}}{\prod_{0 \leq i, j < N} (1 - q^{i + j + 1})}, Δ_{N} = \prod_{d = 1}^{N - 1} (1 - q^{d})^{2 (N - d)} . \end{matrix}

The denominator has limiting multiplicity $d$ at $1 - q^{d}$ , while

\frac{Δ_{N}}{P^{2 N}} = \prod_{d < N} (1 - q^{d})^{- 2 d} \prod_{d \geq N} (1 - q^{d})^{- 2 N} ⟶ M (q)^{2} .

The convergence follows from $\sum d | \log (1 - q^{d}) | < \infty$ and the bound $O_{q} (N q^{N})$ for the logarithm of the second product. Hence

\begin{matrix} (16) & D_{N}^{(0)} \sim M (q)^{3} q^{B_{N}} P^{2 N} . \end{matrix}

Two copies of $M (q)$ come from the squared Vandermonde and one from the Cauchy denominator. We shall compare general weights with this exact model. Fixed displacements of the last few node indices have weight ratios tending to $1$ ; the polynomial bound in the theorem controls the sum over all displacements.

Theorem 3.6 (determinants of geometric moments). Let $a_{k} > 0$ satisfy, for fixed constants $C$ and $κ$ ,

\frac{a_{k + h}}{a_{k}} ⟶ 1 (k \to \infty) for each fixed h \geq 0, \frac{a_{k + h}}{a_{k}} \leq C (1 + h)^{κ} (k, h \geq 0) .

Put $M_{m} = \sum_{k \geq 0} a_{k} q^{(m + 1) k}$ and $D_{N} = det (M_{i + j})_{0 \leq i, j < N}$ . Then

D_{N} \sim M (q)^{3} q^{B_{N}} P^{2 N} \prod_{k = 0}^{N - 1} a_{k} (N \to \infty) .

Proof of Theorem 3.6. We may increase $C$ to at least $1$ and $κ$ to at least $0$ . The bound $a_{h} \leq C a_{0} (1 + h)^{κ}$ makes $\sum_{k \geq 0} a_{k} q^{k} δ_{q^{k}}$ a finite positive measure with moments $M_{m}$ . Apply Cauchy–Binet to finite truncations. Entrywise convergence gives convergence of their determinants, and monotone convergence applies to the nonnegative tuple sums. Hence

\begin{matrix} (17) & D_{N} = \sum_{0 \leq k_{0} < \dots < k_{N - 1}} \prod_{i < N} a_{k_{i}} q^{k_{i}} \prod_{i < j < N} (q^{k_{i}} - q^{k_{j}})^{2} . \end{matrix}

The tuple $k_{i} = i$ contributes $q^{B_{N}} Δ_{N} \prod_{i < N} a_{i}$ . It does not account for the whole asymptotic. For example, replace $(0, 1, \dots, N - 2, N - 1)$ by $(0, 1, \dots, N - 2, N)$ . The ratio of its contribution to that of the original tuple is

\begin{matrix} (18) & q \frac{a_{N}}{a_{N - 1}} \prod_{d = 1}^{N - 1} {(\frac{1 - q^{d + 1}}{1 - q^{d}})}^{2} = q \frac{a_{N}}{a_{N - 1}} {(\frac{1 - q^{N}}{1 - q})}^{2} ⟶ \frac{q}{(1 - q)^{2}} . \end{matrix}

Only pairs involving the moved node change, so the product telescopes. At $q = 1 / 2$ the limiting ratio is $2$ : the displaced tuple contributes twice as much asymptotically as $(0, 1, \dots, N - 1)$ .

Subtracting $(0, 1, \dots, N - 1)$ gives the shifts $λ_{i} = k_{i} - i$ . Since $k_{i + 1} \geq k_{i} + 1$ , these are nonnegative and weakly increasing; reversing their order gives a partition. For $N = 4$ , for example,

\underset{(k_{i})}{\underset{⏟}{(0, 1, 3, 5)}} ⟼ \underset{(λ_{i})}{\underset{⏟}{(0, 0, 1, 2)}} ⟼ \underset{(μ_{j})}{\underset{⏟}{(2, 1, 0, 0)}} .

In general, put $μ_{j} = λ_{N - j}$ for $1 \leq j \leq N$ , so that $μ_{1} \geq \dots \geq μ_{N} \geq 0$ . Division by the contribution of $k_{i} = i$ turns the summand for this tuple into

\begin{matrix} (19) & \begin{aligned} W_{N} (μ) = & q^{\sum_{j \leq N} (2 j - 1) μ_{j}} \prod_{j \leq N} \frac{a_{N - j + μ_{j}}}{a_{N - j}} \\ \cdot \prod_{1 \leq j < k \leq N} {(\frac{1 - q^{k - j + μ_{j} - μ_{k}}}{1 - q^{k - j}})}^{2} . \end{aligned} \end{matrix}

We order the indices increasingly, so their geometric nodes decrease towards zero. The $j$ th index from the right contributes once through $q^{k_{i}}$ and twice through each of its $j - 1$ pairs with later indices in the squared Vandermonde. Its displacement therefore has exponent $1 + 2 (j - 1) = 2 j - 1$ .

Set $W_{N} (μ) = 0$ when $μ$ has more than $N$ positive parts, so that every sum has the same index set. Fix a partition with $ℓ$ positive parts. It affects only the last $ℓ$ indices of $(0, 1, \dots, N - 1)$ : for instance, $(2, 1)$ affects the last two for every $N \geq 2$ . Those indices tend to infinity, so the weight ratios in (19) tend to $1$ by the fixed-shift hypothesis.

The Vandermonde quotient still contains a growing number of factors, but its dependence on $N$ telescopes. The factors with $j < k \leq ℓ$ are fixed, and those with $j > ℓ$ equal $1$ . For $N \geq ℓ$ and each $j \leq ℓ$ , the remaining factors satisfy

\begin{matrix} (20) & \prod_{k = ℓ + 1}^{N} \frac{1 - q^{k - j + μ_{j}}}{1 - q^{k - j}} = \frac{(q^{N + 1 - j}; q)_{μ_{j}}}{(q^{ℓ + 1 - j}; q)_{μ_{j}}} ⟶ \frac{1}{(q^{ℓ + 1 - j}; q)_{μ_{j}}} . \end{matrix}

Thus each $W_{N} (μ)$ has the same limit as for constant weights.

To pass this limit through the sum over partitions, we need a bound independent of $N$ that remains summable as both the number and size of the parts grow. Only factors with $j \leq ℓ$ can differ from $1$ . For each such $j$ , the denominator product in the Vandermonde quotient is at least $P$ , and its numerator product is at most $1$ . Together with the polynomial shift bound, this gives

W_{N} (μ) \leq (C P^{- 2})^{ℓ} \prod_{j = 1}^{ℓ} (1 + μ_{j})^{κ} q^{(2 j - 1) μ_{j}} .

Let $A = \sum_{u \geq 1} (1 + u)^{κ} q^{u - 1} < \infty$ . Dropping the ordering of the positive parts, we obtain

\begin{aligned} \sum_{ℓ (μ) = ℓ} sup_{N} W_{N} (μ) & \leq (C P^{- 2})^{ℓ} \prod_{j = 1}^{ℓ} \sum_{u \geq 1} (1 + u)^{κ} q^{(2 j - 1) u} \\ \leq (C P^{- 2} A)^{ℓ} q^{ℓ^{2}} . \end{aligned}

Here $q^{(2 j - 1) u} \leq q^{2 j - 1} q^{u - 1}$ and $1 + 3 + \dots + (2 ℓ - 1) = ℓ^{2}$ . Since $q < 1$ , the bound $(C P^{- 2} A)^{ℓ} q^{ℓ^{2}}$ is summable over $ℓ$ , independently of $N$ .

Dominated convergence gives the same limiting partition sum as for $a_{k} = 1$ . By (15), this sum tends to $M (q)$ , the reciprocal of the limiting Cauchy denominator. Together with $Δ_{N} / P^{2 N} \to M (q)^{2}$ , it accounts for all three copies of $M (q)$ . Restoring the principal tuple gives

\frac{D_{N}}{D_{N}^{(0)} \prod_{k < N} a_{k}} ⟶ 1.

The model asymptotic (16) now proves the theorem. The comparison explains why the normalised limit retains the geometric spacing of the nodes but loses the individual weights. ◻

We now return to $a_{k} = P^{4} γ_{k}$ . A factorisation into two finite Gaussian multinomial sums supplies both the shift hypotheses and the first correction to $a_{k} / c_{k}$ . The shift hypotheses allow application of the determinant theorem; the correction determines the product $\prod_{k < N} a_{k}$ .

Proposition 3.7 (the moment weights as a product of two finite sums). For $r \geq 1$ and $k \geq 0$ set

R_{k}^{(r)} (q) = \sum_{n_{1} + \dots + n_{r} = k} \frac{(q; q)_{k}}{\prod_{j \leq r} (q; q)_{n_{j}}},

the sum being over compositions into $r$ nonnegative parts, so that $R_{k}^{(r)}$ is the sum of all $q$ -multinomial coefficients of degree $k$ and length $r$ , equivalently the multivariate Rogers–Szegő polynomial at unit arguments [42]. These are polynomial identities in the indeterminate $q$ ; at a prime power, $r = 2$ gives the Galois number counting the subspaces of a $k$ -dimensional space over the field of that order, and $r = 3$ counts the flags $V_{1} \subseteq V_{2}$ in that space. The source is cited for those identifications; the factorisation below is proved here. Then

γ_{k} (q) = \frac{R_{k}^{(2)} (q) R_{k}^{(3)} (q)}{(q; q)_{k}} .

The product $R_{k}^{(2)} R_{k}^{(3)}$ lies in $Z_{\geq 0} [q]$ , has degree $⌊ k^{2} / 4 ⌋ + ⌊ k^{2} / 3 ⌋$ , and has coefficient sum $6^{k}$ . Consequently $(q; q)_{k} γ_{k} \in Z [q]$ , and the weights $a_{k} = P^{4} γ_{k}$ satisfy

P^{5} c_{k} \leq a_{k} \leq P^{- 1} c_{k} (k \geq 0),

together with $a_{k + h} / a_{k} \leq P^{- 6} (1 + h)^{3}$ , so the hypotheses of Theorem 3.6 hold with $κ = 3$ .

Proof. Write $E (z) = (z; q)_{\infty}^{- 1}$ . Euler’s expansion gives $E (z)^{r} = \sum_{n \geq 0} R_{n}^{(r)} z^{n} / (q; q)_{n}$ . We seek the product $R_{k}^{(2)} R_{k}^{(3)}$ at a common index $k$ . Splitting $k = n + j$ keeps $R_{k}^{(3)}$ fixed while the sum over $((q; q)_{n} (q; q)_{j})^{- 1}$ supplies $R_{k}^{(2)} / (q; q)_{k}$ . The operator $\partial_{q} f (z) = (f (z) - f (q z)) / z$ produces the needed index shift: $\partial_{q} z^{j} = (1 - q^{j}) z^{j - 1}$ cancels the last factor of $(q; q)_{j}$ . Thus

\partial_{q} E = E, \partial_{q}^{n} E (z)^{3} = \sum_{j \geq 0} \frac{R_{n + j}^{(3)}}{(q; q)_{j}} z^{j} .

The $q$ -Leibniz rule, obtained by induction from the product rule, is

\partial_{q}^{n} (f g) (z) = \sum_{j = 0}^{n} {[\binom{n}{j}]}_{q} (\partial_{q}^{j} f) (q^{n - j} z) (\partial_{q}^{n - j} g) (z) .

Write $h_{k} (z) = \sum_{j = 0}^{k} {[\binom{k}{j}]}_{q} (z; q)_{j}$ . Applying the rule to $E \cdot E$ and using $E (q^{m} z) = (z; q)_{m} E (z)$ gives $\partial_{q}^{k} E (z)^{2} = E (z)^{2} h_{k} (z)$ ; at $k = 1$ both sides are $E (z)^{2} (2 - z)$ . Applying it once more, to $E \cdot E^{2}$ , gives

\partial_{q}^{n} E (z)^{3} = E (z)^{3} \sum_{k = 0}^{n} {[\binom{n}{k}]}_{q} (z; q)_{k} h_{k} (z),

Hence

\sum_{n \geq 0} \frac{w^{n}}{(q; q)_{n}} \partial_{q}^{n} E (z)^{3} = E (w) E (z)^{3} \sum_{k \geq 0} \frac{w^{k} (z; q)_{k}}{(q; q)_{k}} h_{k} (z) .

The left-hand side is $\sum_{n, j \geq 0} R_{n + j}^{(3)} w^{n} z^{j} / ((q; q)_{n} (q; q)_{j})$ . On setting $z = w$ , all pairs with the same sum contribute to a single coefficient. Thus the coefficient of $w^{k}$ is $R_{k}^{(3)} \sum_{j \leq k} ((q; q)_{j} (q; q)_{k - j})^{- 1} = R_{k}^{(2)} R_{k}^{(3)} / (q; q)_{k}$ . On the right-hand side, put $z = w$ , expand $h_{k}$ , and write $k = j + t$ . The identity $(w; q)_{j + t} = (w; q)_{j} (w q^{j}; q)_{t}$ gives

E (w)^{4} \sum_{j \geq 0} \frac{w^{j} (w; q)_{j}^{2}}{(q; q)_{j}} \sum_{t \geq 0} \frac{(w q^{j}; q)_{t}}{(q; q)_{t}} w^{t} .

The inner sum is evaluated by the $q$ -binomial theorem:

\sum_{t \geq 0} \frac{(w q^{j}; q)_{t}}{(q; q)_{t}} w^{t} = \frac{(w^{2} q^{j}; q)_{\infty}}{(w; q)_{\infty}} .

Finally, $(w; q)_{j} = (w; q)_{\infty} / (w q^{j}; q)_{\infty}$ cancels two of the five factors $E (w)$ , leaving

E (w)^{3} \sum_{j \geq 0} \frac{w^{j}}{(q; q)_{j}} \frac{(w^{2} q^{j}; q)_{\infty}}{(w q^{j}; q)_{\infty}^{2}} = G_{q} (w) .

Every step holds coefficientwise, and absolutely for $| w |$ small.

A $q$ -multinomial coefficient has nonnegative integral coefficients and degree $\sum_{i < j} n_{i} n_{j}$ . At $q = 1$ it equals the corresponding ordinary multinomial coefficient. Summing over compositions gives coefficient sum $r^{k}$ for $R_{k}^{(r)}$ and hence $6^{k}$ for the product. The degree is maximised by balanced parts, which gives $⌊ k^{2} / 4 ⌋$ for $r = 2$ and $⌊ k^{2} / 3 ⌋$ for $r = 3$ ; no leading cancellation occurs because all coefficients are nonnegative.

For the bounds, write $b_{k}^{(r)} = [z^{k}] (z; q)_{\infty}^{- r}$ , so that $R_{k}^{(r)} = (q; q)_{k} b_{k}^{(r)}$ and $γ_{k} = (q; q)_{k} b_{k}^{(2)} b_{k}^{(3)}$ . Euler’s expansion exhibits $b_{k}^{(r)}$ as an $r$ -fold convolution of the sequence $1 / (q; q)_{n}$ , whose terms lie in $[1, P^{- 1}]$ ; the number of compositions of $k$ into $r$ nonnegative parts is $(\binom{k + r - 1}{r - 1})$ , so

(\binom{k + r - 1}{r - 1}) \leq b_{k}^{(r)} \leq P^{- r} (\binom{k + r - 1}{r - 1}) .

Combining these at $r = 2, 3$ with $P \leq (q; q)_{k} \leq 1$ and $c_{k} = (k + 1) (\binom{k + 2}{2})$ gives $P^{5} c_{k} \leq a_{k} \leq P^{- 1} c_{k}$ . The shift bound follows from $c_{k + h} / c_{k} \leq (1 + h)^{3}$ .

It remains to show that $a_{k + h} / a_{k} \to 1$ for each fixed $h$ . The two convolutions grow at least linearly and quadratically; it suffices to bound their increments by a constant and a linear function, respectively. Put $e_{n} = 1 / (q; q)_{n}$ . Since $(q; q)_{n + 1} = (1 - q^{n + 1}) (q; q)_{n}$ , these terms satisfy $1 = e_{0} \leq e_{1} \leq \dots \leq P^{- 1}$ , and the convolutions read $b_{k}^{(2)} = \sum_{i \leq k} e_{i} e_{k - i}$ and $b_{k}^{(3)} = \sum_{i \leq k} e_{i} b_{k - i}^{(2)}$ . Hence

\begin{aligned} b_{k + 1}^{(2)} - b_{k}^{(2)} & = \sum_{i \leq k} e_{i} (e_{k + 1 - i} - e_{k - i}) + e_{k + 1}, \\ b_{k + 1}^{(3)} - b_{k}^{(3)} & = \sum_{i \leq k} e_{i} (b_{k + 1 - i}^{(2)} - b_{k - i}^{(2)}) + e_{k + 1} . \end{aligned}

Every bracket is nonnegative. In the first line the brackets add up to $e_{k + 1} - 1 < P^{- 1}$ and each $e_{i} \leq P^{- 1}$ , so $0 \leq b_{k + 1}^{(2)} - b_{k}^{(2)} \leq P^{- 2} + P^{- 1} \leq 2 P^{- 2}$ . Inserting this bound into the $k + 1$ terms of the second line gives $0 \leq b_{k + 1}^{(3)} - b_{k}^{(3)} \leq 2 (k + 1) P^{- 3} + P^{- 1} \leq 3 (k + 1) P^{- 3}$ . Against the lower bounds $b_{k}^{(2)} \geq k + 1$ and $b_{k}^{(3)} \geq (k + 1) (k + 2) / 2$ displayed above, the ratio $b_{k + 1}^{(r)} / b_{k}^{(r)}$ lies in $[1, 1 + 2 P^{- 2} / (k + 1)]$ for $r = 2$ and in $[1, 1 + 6 P^{- 3} / (k + 2)]$ for $r = 3$ . Both ratios tend to $1$ , and therefore so does $b_{k + h}^{(r)} / b_{k}^{(r)}$ , a product of $h$ consecutive ratios. Finally $(q; q)_{k + h} / (q; q)_{k} = \prod_{j = 1}^{h} (1 - q^{k + j}) \to 1$ , and $a_{k + h} / a_{k} = γ_{k + h} / γ_{k}$ is the product of these three ratios. The Lean proof obtains the fixed-shift limit by the same steps. ◻

The factorisation proves positivity at each fixed real $q$ , not coefficientwise positivity in $q$ : for example, $a_{0} = P^{4}$ has $q$ -coefficient $- 4$ . The cubic factors $c_{k}$ multiply to $C_{N}$ , but $a_{k} / c_{k} \to 1$ alone would not determine the product of $N$ weights. The first relative correction has order $1 / k$ and accumulates as a harmonic sum; the remaining $O_{q} (k^{- 2})$ terms determine a finite constant. The two convolutions supply this first correction explicitly.

Theorem 3.8 (the size of $V_{N}^{*}$ at a fixed base). Let $γ_{E}$ be Euler’s constant. The product

A (q) = e^{- 8 γ_{E} L} \prod_{k \geq 0} (\frac{a_{k}}{c_{k}} e^{8 L / (k + 1)})

converges to a positive finite value. With $K (q) = A (q) M (q)^{3}$ ,

V_{N}^{*} (q) \sim K (q) C_{N} q^{B_{N}} P^{2 N} N^{- 8 L},

equivalently

\log V_{N}^{*} (q) = B_{N} \log q + \log C_{N} + 2 N \log P - 8 L \log N + \log K (q) + o (1) .

Proof of Theorem 3.8. Put $L = F (1 / q)$ , $e_{k} = (q; q)_{k}^{- 1}$ and $E = P^{- 1}$ . Replacing every $e_{k}$ by $E$ in the two convolution sums gives $E^{2} (k + 1)$ and $E^{3} (\binom{k + 2}{2})$ . The factors $k + 1$ and $(\binom{k + 2}{2})$ count the nonnegative pairs and triples with sum $k$ , respectively. For a fixed index $i$ , there is one pair $(i, k - i)$ and there are $k - i + 1$ triples whose first coordinate is $i$ . Thus a summable deficit in the sequence $e_{i}$ produces a constant correction in the two-fold sum and a linear correction in the three-fold sum. We calculate that deficit and bound terms with two or more deficits.

Writing $d_{k} = E - e_{k}$ , the identity $e_{k + 1} - e_{k} = q^{k + 1} e_{k + 1}$ gives

d_{k} = \sum_{j > k} q^{j} e_{j}, 0 \leq d_{k} \leq \frac{E q^{k + 1}}{1 - q} .

Thus $\sum d_{k}$ and $\sum k d_{k}$ converge. To evaluate the former, use Euler’s identity $E (z) = \sum_{k \geq 0} e_{k} z^{k} = (z; q)_{\infty}^{- 1}$ for $| z | < 1$ . Tonelli’s theorem and logarithmic differentiation at $z = q$ give

S := \sum_{k \geq 0} d_{k} = \sum_{j \geq 1} j q^{j} e_{j} = q E^{'} (q) = E \sum_{r \geq 1} \frac{q^{r}}{1 - q^{r}} = E L .

In the double sum, $q^{j} e_{j}$ occurs once for each $k = 0, \dots, j - 1$ , which explains the factor $j$ . Logarithmic differentiation is justified by locally uniform convergence on $| z | < 1$ .

Let $b_{k}^{(r)} = [z^{k}] E (z)^{r}$ . The finite-sum formula in Proposition 3.7 becomes

a_{k} = P^{4} (q; q)_{k} b_{k}^{(2)} b_{k}^{(3)} .

We expand each finite convolution using $e_{k} = E - d_{k}$ . For two factors,

b_{k}^{(2)} = E^{2} (k + 1) - 2 E \sum_{i = 0}^{k} d_{i} + \sum_{i = 0}^{k} d_{i} d_{k - i} = E^{2} (k + 1 - 2 L) + O_{q} ((k + 1) q^{k}) .

For three factors, the terms with one $d_{i}$ contribute $- 3 E^{2} \sum_{i = 0}^{k} (k - i + 1) d_{i}$ . Here the sum equals $(k + 1) S + O_{q} (1)$ because $\sum i d_{i} < \infty$ . Terms with two $d_{i}$ are bounded by $3 E S^{2}$ , and the term with three is $O_{q} ((k + 1)^{2} q^{k})$ . It follows that

b_{k}^{(3)} = E^{3} ((\binom{k + 2}{2}) - 3 L (k + 1)) + O_{q} (1) .

Dividing the correction $- 3 L (k + 1)$ by the leading term $(\binom{k + 2}{2})$ gives the relative correction $- 6 L / (k + 2)$ . The two-fold convolution contributes $- 2 L / (k + 1)$ in the same way. Since $(q; q)_{k} = P (1 + O_{q} (q^{k}))$ , the leading constants in $a_{k} = P^{4} (q; q)_{k} b_{k}^{(2)} b_{k}^{(3)}$ multiply to $P^{5} E^{5} = 1$ . Replacing $(k + 2)^{- 1}$ by $(k + 1)^{- 1}$ changes only the $O_{q} ((k + 1)^{- 2})$ error. Thus the coefficients $2 L$ and $6 L$ add, and with $c_{k} = (k + 1)^{2} (k + 2) / 2$ we obtain

\begin{aligned} \frac{a_{k}}{c_{k}} & = (1 - \frac{2 L}{k + 1} + O_{q} (q^{k})) (1 - \frac{6 L}{k + 2} + O_{q} ((k + 1)^{- 2})) (1 + O_{q} (q^{k})) \\ = 1 - \frac{8 L}{k + 1} + O_{q} ((k + 1)^{- 2}) . \end{aligned}

In particular $a_{k + 1} / a_{k} \to 1$ , and multiplication of finitely many adjacent ratios gives $a_{k + h} / a_{k} \to 1$ for each fixed $h$ . The uniform shift bound follows from Proposition 3.7, so Theorem 3.6 applies.

For large $k$ , taking logarithms preserves the first-order term and leaves the summable error $\log (a_{k} / c_{k}) + 8 L / (k + 1) = O_{q} ((k + 1)^{- 2})$ . The finitely many earlier factors are positive, so the regularised product also has a positive finite limit: $0 < A (q) < \infty$ . Since $\prod_{k < N} c_{k} = C_{N}$ , the harmonic contribution separates as

\log \frac{\prod_{k < N} a_{k}}{C_{N}} = - 8 L \sum_{k < N} \frac{1}{k + 1} + \sum_{k < N} (\log \frac{a_{k}}{c_{k}} + \frac{8 L}{k + 1}) .

The second sum tends to $\log A (q) + 8 γ_{E} L$ . Using $\sum_{k < N} (k + 1)^{- 1} = \log N + γ_{E} + o (1)$ , the Euler-constant terms cancel, giving $\prod_{k < N} a_{k} \sim A (q) C_{N} N^{- 8 L}$ . Theorem 3.6 now gives the stated determinant asymptotic. Taking logarithms yields

\log V_{N}^{*} = B_{N} \log q + \log C_{N} + 2 N \log P - 8 L \log N + \log K (q) + o (1) .

◻

The convolution proof explains $8 L$ as $2 L + 6 L$ , one contribution from each finite sum. An alternative reads the same correction from the generating function near $w = 1$ . Its pole of order four accounts for the cubic growth of $a_{k}$ ; the next principal-part coefficient accounts for the relative term of order $1 / k$ .

An alternative coefficient calculation. Write $Π (w) = (q w; q)_{\infty}$ , so that $(w; q)_{\infty} = (1 - w) Π (w)$ , and for $t \geq 1$ put

f_{t} (w) = \frac{(q^{t} w^{2}; q)_{\infty}}{(q; q)_{t} (q^{t} w; q)_{\infty}^{2}}, S (w) = \sum_{t \geq 1} w^{t} (f_{t} (w) - P^{- 1}) .

On each compact subset of $| w | < q^{- 1}$ the bracket is $O_{q} (q^{t})$ uniformly, so $S$ is analytic there. Separating the summand $t = 0$ , whose denominator carries the factor $(1 - w)^{2}$ , and subtracting the constant part of the remaining summands gives

D (w) := P^{4} (1 - w)^{4} G_{q} (w) = \frac{P^{4}}{Π (w)^{3}} ((1 + w) \frac{(q w^{2}; q)_{\infty}}{Π (w)^{2}} + \frac{w}{P} + (1 - w) S (w)) .

Hence $D$ is analytic on a disc of radius greater than $1$ .

At $w = 1$ one has $Π (1) = P$ and $Π^{'} (1) / Π (1) = - L$ , since $Π^{'} / Π = - \sum_{j \geq 1} q^{j} / (1 - q^{j} w)$ . Also $(q w^{2}; q)_{\infty}$ equals $P$ at $w = 1$ , and the logarithmic derivative of $(q w^{2}; q)_{\infty} / Π (w)^{2}$ at $w = 1$ is $- 2 L + 2 L = 0$ . Therefore the bracket has value $3 / P$ and derivative $(2 - L) / P$ at $w = 1$ ; here $S (1) = L / P$ , because $f_{t} (1) = (P (1 - q^{t}))^{- 1}$ and $\sum_{t \geq 1} ((1 - q^{t})^{- 1} - 1) = L$ . Combining with the factor $P^{4} Π^{- 3}$ , whose logarithmic derivative at $1$ is $3 L$ , gives

D (1) = 3, D^{'} (1) = 3 (3 L + \frac{2 - L}{3}) = 2 + 8 L .

The first two principal-part terms are determined by $D (1)$ and $D^{'} (1)$ . The terms from its second and third derivatives have coefficients of order at most $k$ ; after the full principal part is subtracted, the remainder is analytic on a larger disc and has exponentially decreasing coefficients. Keeping only the first two terms therefore gives

a_{k} = 3 (\binom{k + 3}{3}) - (2 + 8 L) (\binom{k + 2}{2}) + O_{q} (k), \frac{a_{k}}{c_{k}} = 1 - \frac{8 L}{k + 1} + O_{q} ((k + 1)^{- 2}),

the second line because $3 (\binom{k + 3}{3}) = c_{k} (k + 3) / (k + 1)$ and $(\binom{k + 2}{2}) = c_{k} / (k + 1)$ .

The logarithmic error is again summable. The product argument in the first proof therefore gives $0 < A (q) < \infty$ and, using $\sum_{k < N} (k + 1)^{- 1} = \log N + γ_{E} + o (1)$ ,

\prod_{k < N} a_{k} \sim A (q) C_{N} N^{- 8 L} .

Proposition 3.7 supplies the uniform shift bound and the fixed-shift limit, so Theorem 3.6 applies to $(a_{k})$ and gives $V_{N}^{*} \sim M (q)^{3} q^{B_{N}} P^{2 N} \prod_{k < N} a_{k}$ . Substituting the product asymptotic proves both displays. ◻

Comparison with the cubic bound.

An extension with limiting weight ratio

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The size of 𝑉∗𝑁VN∗ at a fixed base

About this paper

The size of $V_{N}^{*}$ at a fixed base