$$ n_k = [\frac{(k-1)(1-\rho)}{1 + (1-\rho)k}][\frac{(k-2)(1-\rho)}{1 + (1-\rho)(k-1)}]... $$

Where $k \geq 1$ and $0 < \rho < 1$. My interpretation is:

$$ \frac{(1-\rho)^{k-1}}{(1-\rho)^k}[\frac{(k-1)}{(\frac{1}{1-\rho}+k)}][\frac{(k-2)}{(\frac{1}{1-\rho}+(k-1))}]... $$

But I think I am missing something.


Let's assume $\rho \neq 1$ and $k\geq 2$. Each factor of $n_k$ has the shape \begin{align*} \frac{(j-1)(1-\rho)}{1+(1-\rho)j}=\frac{1-\rho}{1-\rho}\cdot \frac{j-1}{\frac{1}{1-\rho}+j}=\frac{j-1}{\frac{1}{1-\rho}+j}\qquad\qquad 2\leq j \leq k \end{align*}

We obtain using the product symbol \begin{align*} n_k=\prod_{j=2}^k\frac{(j-1)(1-\rho)}{1+(1-\rho)j}=\prod_{j=2}^k\frac{j-1}{\frac{1}{1-\rho}+j} =(k-1)!\prod_{j=2}^k\frac{1}{\frac{1}{1-\rho}+j} \end{align*}

Hint: The problem is not clearly specified. To fully state the problem we need to know assertions like $\rho \neq 1$ and ranges of variables ($k\geq 2$). We should also specify that the rightmost factor of $n_k$ is $\frac{1\cdot(1-\rho)}{1+(1-\rho)\cdot2}$ to indicate that we don't want a factor $0$ to be part of this product.


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