\ldots n_{d-k-1}!} \exp (\beta e^{\nu_{d-k} b^k q}),\end{gathered}$$ where $\alpha_{p} = \exp ( \beta e^{\nu_{d-k} b^k p}k ) $. Again we changed $(q+1)^{d-k-1}$ terms of infinite series and estimated exponential part of the series by $\exp(\beta e^{\nu_{d-k}q})$. Both finite series tend to zero when divided by $e^{\beta (d-k-1)}$ and the infinite series is equal to $$\begin{gathered}
\nonumber \exp ( \beta e^{\nu_{d-k} b^k p}k +\beta e^{\nu_{d-k} b^k}(d-k-1) + \beta e^{\nu_{d-k} b^k q}).\end{gathered}$$ We have to select $p$ and $q$ so that $$\begin{gathered}
\nonumber R_1=e^{\nu_{d-k} b^k p}k +e^{\nu_{d-k} b^k}(d-k-1) + e^{\nu_{d-k} b^k q} - (d-k-1)<0,\end{gathered}$$ which is possible for any $\nu_{d-k}<0.$ Then we reverse the substitution $\beta=\frac{aT}{d}$ and the ratio $\frac{A(a)}{B(a)}$ tends to zero for $a\rightarrow\infty.$ Its convergence is not slower than $e^{Ra},$ where $R=\frac{TR_1}{d}.$ $\Box $

\[tecko\]For $a\rightarrow\infty$ it holds $$\label{a}{\mathbb E} G_{d-k}(\mu^{(d-k)}_a)\rightarrow 0,\; k=0,\dots ,d-2.$$

[**Proof:**]{} In (\[znovu\]) it suffices to show that for $y_1,\dots ,y_{d-k}$ all with distinct orientation we have $$\label{roj}\lim_{a\rightarrow\infty}a^{d-k}\rho_{d-k}(y_1,\dots ,y_{d-k};\mu_a^{(d-k)})=0.$$ This is the consequence of Lemmas \[ll5\] and \[ll6\]. Then by the Lebesgue dominance theorem formula (\[a\]) follows. $\Box $

Since the functionals $G_{d-k}$ are nonnegative Theorem \[tecko\] says that asymptotically the processes $\mu_a^{d-k}$ are degenerate (for any $\nu_{d-k}<0$) in the sense that there are no intersections 