\frac{e(1-e^{-t}) r^{nt} \mbox{vol }\mathcal{B}_1}{nt}.$$ Choose $r$ such that the right-hand side of this last equation is equal to $r(K)V\zeta(nt)(1-\varepsilon)/n$ for a small $\varepsilon < 1$. Let $\phi_p$ be as in Definition \[def:ConsGen\] and $\mathbb{L}_p$ its induced ensemble $$\mathbb{L}_p=\left\{ \beta \Lambda_p(C) : C \mbox{ is an } (n,k,p)-\mbox{code}\right\}$$ as in Equation . According to Theorem \[thm:AverageEnsembleGeneral\], it is possible to find $\Lambda_1 = \beta \sigma(\Lambda_0) \in \mathbb{L}_{p}$ of volume $V$ such that, for $p$ sufficiently large $$\sum_{{\mathbf{y}}\in\Lambda_1^{\prime}} f({\mathbf{y}}) \leq (1-\varepsilon) \frac{r(K)}{n} < \frac{r(K)}{n},$$ Let $v_1, \ldots, v_t$ be linearly independent vectors in $\Lambda_0$ achieving the successive minima, $\left\|\beta \sigma(v_i)\right\|=\lambda_i^K(\Lambda_0)$. We have $$\sum_{{\mathbf{y}}\in\Lambda_1^{\prime}} f({\mathbf{y}}) \geq \sum_{i=1}^t \sum_{\mu} f(\beta \sigma(\mu v_i))  = r(K) \sum_{i=1}^t f(\beta \sigma(v_i)),$$ where the sum with subscript $\mu$ is over all roots of unity in $K$. From this we conclude that, for all $i$, $\beta \lambda_i^K(\Lambda_0) \geq r e^{1/n-1}$ and $$\frac{1}{n} - \log\left(\frac{\beta^t 
\prod \lambda_i^K(\Lambda_0)}{r^t}\right) < \frac{1}{n}.$$ Therefore, for the $t$ successive densities (Eq. ):

$$\left(\prod_{i=1}^{t} \Delta_i^K\right)^{1/t} = \prod_{i=1}^t \left( \frac{\mbox{vol}(\mathcal{B}_{\lambda_i^K/2})}{V} \right)^{1/t} \geq \frac{ r(K) t \zeta(nt) (1-\varepsilon)}{e(1-e^{-t})2^{nt}}.$$

But in this case, we can find $\tilde{\Lambda}$ whose packing density (or $\Delta_1^K$) is greater or equal than $\frac{ r(K) t \zeta(nt)}{e(1-e^{-t})2^{nt}(1-\varepsilon)}$ (e.g. [@Vance Thm. 2.2]).

Balanced Sets of Codes over 