\subset [d-k]} \prod_{j=1}^{d-k}n_{l_{j}}) = \\
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-k}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d-k}}}{n_{1}! \ldots n_{d-k}!} \exp( \nu_{d-k} (2b)^k  \prod_{i=1}^{d-k}n_{i}) =\\
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-k-1}=0}^{\infty}\frac{\beta^{n_{1}+\ldots + n_{d-k-1}}}{n_{1}! \ldots n_{d-k-1}!} \exp\left( \beta e^{\nu_{d-k} (2b)^{k}  \prod_{i=1}^{d-k-1}n_{i} } \right) \geq \\
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-k-1}=0}^{\infty}\frac{\beta^{n_{1}+\ldots + n_{d-k-1}}}{n_{1}! \ldots n_{d-k-1}!} = e^{\beta(d-k-1)} . \end{gathered}$$$\Box $

\[ll6\] There exists a constant $R<0$ independent of $a$ such that $$\label{oddh} \frac{A(a)}{B(a)}\leq e^{Ra}$$

[**Proof:**]{} Substituting $\beta=\frac{aT}{d}$ we estimate $A(a)$ $$\label{prsu}
A(a)\leq\sum_{n=0}^{\infty}\beta^{n}\sum_{n_{1}+ \ldots +n_{d} = n} \frac{1}{n_{1}! \ldots n_{d}!} \exp ( \nu_{d-k} b^k C_{d-k,d}),$$ where $$\begin{aligned}
\nonumber C_{l,m} &= \sum_{i=1}^{l} \sum_{\{j_{1}, \ldots , j_{i}\} \subset [m]} \prod_{k=1}^{i} n_{j_{k}}+1,\;1\leq l\leq m. \\  
\nonumber C_{l,m} &= C_{l-1,m-1}n_{m}+C_{l,m-1},\;2\leq l<m, \\
\nonumber C_{m,m} &= C_{m-1,m-1}(1+n_{m}).\end{aligned}$$ Now we decrease the number of sums in (\[prsu\]): $$\begin{gathered}
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d}}}{n_{1}! \ldots n_{d}!} \exp ( \nu_{d-k} b^k C_{d-k,d}) = \\
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d}}}{n_{1}! \ldots n_{d}!} \exp ( \nu_{d-k} b^k (C_{d-k-1,d-1}n_{d}+C_{d-k,d-1} ))= \\
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-1}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d-1}}}{n_{1}! \ldots n_{d-1}!} \exp (\beta e^{\nu_{d-k} b^k C_{d-k-1,d-1}}  + \nu_{d-k} b^k C_{d-k,d-1} ) \leq \\
\nonumber \sum_{n_{1}=0}^{\infty} \ldots \sum_{n_{d-1}=0}^{\infty} \frac{\beta^{n_{1}+\ldots + n_{d-1}}}{n_{1}! \ldots n_{d-1}!} \exp ( \beta e^{\nu_{d-k} b^k p}  + \nu_{d-k} b^k C_{d-k,d-1} ) + \\ 
\nonumber \sum_{n_{1}=0}^{p} \ldots \sum_{n_{d-1}=0}^{p} \frac{\beta^{n_{1}+\ldots 