$$R_1 \leq \frac{1}{2}\log\left(\frac{P_1}{\sum_{k=2}^N (2a^2 + 1)P_k + 1}\right).$$ Proceeding along similar lines, at stage $i$, interference $I_{1i}^n$ and message $m_{1i}$ can be decoded successfully if $$\begin{aligned}
R_i \leq \frac{1}{2}\log\left(\frac{a^2 P_i}{P_i + \sum_{k=i+1}^N (2a^2 + 1)P_k + 1}\right),\\
R_i \leq \frac{1}{2}\log\left(\frac{P_i}{\sum_{k=i+1}^N (2a^2 + 1)P_k + 1}\right).\end{aligned}$$ The power values have been chosen so that the “very strong" interference condition is satisfied at each stage. The noise plus interference power seen at stage $i$ in decoding interference $I_{1i}^n$ and message $m_{1i}$ is equal to $\sum_{k=i+1}^N (2a^2 + 1)P_k + 1$. From the power assignments in (\[eqn : power assignment 1\]), we can see that $$a^2 = \frac{P_i}{\sum_{k=i+1}^N (2a^2 + 1)P_k + 1} + 1.$$ With the choice of power values as in (\[eqn : power assignment 1\]), the rate at each stage is given by $$R_i = \frac{1}{2}\log(a^2 - 1).$$ For $R_i$ to be positive, we need $a^2 \geq 2$. Hence, the total rate achieved by each user is given by $$R = \frac{1}{2}\log(a^2-1)^N.$$ Also, the total power used by each transmitter is given by $$\begin{aligned}
P &=& P_1 + \ldots + P_N\nonumber\\
   &\leq & (2a^4-a^2)^N.\end{aligned}$$ Taking $N$ to $\infty$, we get the desired result. That is, $$\lim_{N, P \rightarrow \infty} \frac{3 