simplifying a summation related to an epidemic process Here is a question that I came across while trying to understand some fundamental mathematical concepts related to the spread of epidemics. The problem turns out to be a mathematical puzzle related to the binomial coefficient.
Consider an epidemic process such as the one described in   [zhang2017contact].  Let $\lambda(N)$ be an exogenous infection rate. Let $\gamma^{N/2}$ be an endogenous infection rate.  The total rate at which a node is contaminated (endogenously and exogenously) is the product of these two quantities. Let also $\mu$ be the recovery rate.    The ratio of the contamination rate over the recovery rate plays an  important role in the study of epidemics.
The probability that $i$ node are infected is    $\tilde{\pi}(i;N)/Z$ where
    \begin{equation}
  \tilde{\pi}(i;N)={N \choose i} \left( \frac{\lambda(N)}{\mu} \gamma^{N/2} \right )^{i}  \textrm{ , } i =0,\ldots, N
 \end{equation}
The fraction of infected nodes is
\begin{equation}
 \rho(N) = \frac{E(I)}{N} = \sum_{i=0}^ N i \pi(i;N)/{N}
\end{equation}
where
\begin{equation}
\pi(i;N) = \frac{\tilde{\pi}(i;N)}{Z}
\end{equation}
and
\begin{equation}
Z=\sum_{i=0}^N   \tilde{\pi}(i;N)
\end{equation}
Questions
Our goal is to analyze $\rho(N)$ and to show that, for certain shapes of $\lambda(N)$, we have that $\rho(N)$ first decreases and then increases. This means that, as $N$ increases, the system is first dominated by  exogenous factors, and then by endogenous factors.
1) for which shapes of the $\lambda(N)$ function we are able to obtain simple closed form expressions for $\rho$?
2) this is the case of most interest: are we able to treat the case $\lambda(N)=1/N$?
Zhang et al.,  Contact process with exogenous infection and the scaled SIS process. Journal of Complex Networks, 2017


We obtain
  \begin{align*}
\color{blue}{\rho(n)}&=\frac{1}{N}\sum_{j=0}^N j\pi(j;N)\\
&=\frac{1}{N}\sum_{j=0}^N j\binom{N}{j}\left(\frac{\lambda(N)}{\mu}\gamma^{\frac{N}{2}}\right)^j\\
&=\sum_{j=1}^N\binom{N-1}{j-1}\left(\frac{\lambda(N)}{\mu}\gamma^{\frac{N}{2}}\right)^j\tag{1}\\
&=\sum_{j=0}^{N-1}\binom{N-1}{j}\left(\frac{\lambda(N)}{\mu}\gamma^{\frac{N}{2}}\right)^{j+1}\tag{2}\\
&\color{blue}{=\left(\frac{\lambda(N)}{\mu}\gamma^{\frac{N}{2}}\right)\left(1+\frac{\lambda(N)}{\mu}\gamma^{\frac{N}{2}}\right)^{N-1}}\tag{3}
\end{align*}

Comment:


*

*In (1) we use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.

*In (2) we shift the index to start with $j=0$.

*In (3) we apply the binomial theorem.
