TITLE: Computing a limit of a sum mixed with product.
QUESTION [10 upvotes]: Given $d\ge 2$ integer and  $m_0>0$ define
$$m_k= m_0\left(\frac{d}{d-1}\right)^k\quad \text{and}\quad\sigma_n= \frac{1}{m_n+d}$$
I would like to compute
$$\lim_{n\to \infty}\prod_{j=1}^{n}(1-\sigma_j)$$
and
$$\lim_{n\to \infty}\sum_{k=1}^{n-2} \sigma_k\prod_{j=k+1}^{n-1}(1-\sigma_j)$$
I expect the following results  $\frac{m_0}{m_0+d}$ for the first and  $\frac{1}{m_0+d}$ for the second.

REPLY [4 votes]: We have, $\dfrac{m_k+d-1}{m_{k+1}+d}=\dfrac{d-1}{d}$, $\forall{k}$
$\lim\limits_{n\to \infty}\prod\limits_{j=1}^{n}(1-\sigma_j)=\lim\limits_{n\to \infty}\prod\limits_{j=1}^{n}\dfrac{m_j+d-1}{m_j+d}=\lim\limits_{n\to \infty}\left(\dfrac{1}{m_1+d}\right).\prod\limits_{j=1}^{n-1}\dfrac{m_j+d-1}{m_{j+1}+d}.\left(m_n+d-1\right)$
$=\left(\dfrac{1}{m_1+d}\right).\lim\limits_{n\to \infty}\left(\dfrac{d-1}{d}\right)^{n-1}.\left(m_0.\left(\dfrac{d}{d-1}\right)^n+d-1\right)=\dfrac{m_0.\dfrac{d}{d-1}}{m_0.\dfrac{d}{d-1}+d}=\dfrac{m_0}{m_0+d-1}$
Also, we have,
$t_n^k=\sigma_k\prod\limits_{j=k+1}^{n-1}(1-\sigma_j) = \sigma_k \prod\limits_{j=k+1}^{n-1}\dfrac{m_j+d-1}{m_j+d}$
$= \sigma_k \dfrac{1}{m_{k+1}+d}\prod\limits_{j=k+1}^{n-2}\dfrac{m_j+d-1}{m_{j+1}+d}.(m_{n-1}+d-1)$
$=\sigma_k\sigma_{k+1}\left(\dfrac{d-1}{d}\right)^{n-k-2}.\left(m_0.\left(\dfrac{d}{d-1}\right)^{n-1}+d-1\right)$
$=\sigma_k\sigma_{k+1}\left(m_0.\left(\dfrac{d}{d-1}\right)^{k+1}+d\left(\dfrac{d-1}{d}\right)^{n-k-1}\right)$
$=\sigma_k\sigma_{k+1}\left(m_{k+1}+d\left(\dfrac{d-1}{d}\right)^{n-k-1}\right)$
$\Rightarrow \lim\limits_{n\to \infty} t_{n}^{k} = \sigma_k\sigma_{k+1}m_{k+1}$
We can see  $\{{t_{n}^{k}\}}$ is a decreasing sequence for a fixed $k$ and it converges to $\sigma_k\sigma_{k+1}m_{k+1}$ at $n \to \infty$.
Now, as pointed out by @Wiley, we have $\dfrac{1}{\sigma_{k+1}}-\dfrac{1}{\sigma_k}=(m_{k+1}+d) - (m_k+d)=m_{k+1}-m_k=\left(1-\dfrac{d-1}{d}\right)m_{k+1}=\dfrac{m_{k+1}}{d}$
$\implies \dfrac{\sigma_k-\sigma_{k+1}}{\sigma_k\sigma_{k+1}}=\dfrac{m_{k+1}}{d}$
$\implies \sigma_k\sigma_{k+1}m_{k+1}=d(\sigma_k-\sigma_{k+1})$
$\implies \sum\limits_{k=1}^{n-2}\sigma_k\sigma_{k+1}m_{k+1} = d\sum\limits_{k=1}^{n-2}(\sigma_k-\sigma_{k+1})=d(\sigma_1-\sigma_{n-1})$
$\therefore \lim\limits_{n\to \infty}\sum\limits_{k=1}^{n-2} \sigma_k\prod\limits_{j=k+1}^{n-1}(1-\sigma_j) = \lim\limits_{n\to \infty}\sum\limits_{k=1}^{n-2} t_n^k $
$\geq \lim\limits_{n\to \infty}\sum\limits_{k=1}^{n-2} \lim\limits_{n\to \infty}t_n^k$, since $t_n^k$ decreasing for a fixed $k$
$= \lim\limits_{n\to \infty} \sum\limits_{k=1}^{n-2}\sigma_k\sigma_{k+1}m_{k+1} $
$= d\lim\limits_{n\to \infty}(\sigma_1-\sigma_{n-1})=d\sigma_1$, since $\lim\limits_{n\to \infty} \sigma_{n-1}=0$
$=\dfrac{d}{m_1+d} =\dfrac{d-1}{m_0+d-1}$.
Also, $\sum\limits_{k=1}^{n-2}t_n^k < \sum\limits_{k=1}^{n-2} \sigma_k\sigma_{k+1}m_{k+1} + \sum\limits_{k=1}^{n-2}\sigma_0\sigma_1 d\left(\dfrac{d}{d-1}\right)^{n-k-1}$, since $\{\sigma_k\sigma_{k+1}\}$ is decreasing
$\implies \lim\limits_{n\to\infty}\sum\limits_{k=1}^{n-2}t_n^k < \lim\limits_{n\to\infty}\sum\limits_{k=1}^{n-2} \sigma_k\sigma_{k+1}m_{k+1} + d^2\sigma_0\sigma_1$
$\quad \quad \quad \quad \quad \quad =\dfrac{d-1}{m_0+d-1}+\dfrac{d(d-1)}{(m_0+d)(m_0+d-1)}$
Hence, we have $\dfrac{d-1}{m_0+d-1} \leq \lim\limits_{n\to \infty}\sum\limits_{k=1}^{n-2} \sigma_k\prod\limits_{j=k+1}^{n-1}(1-\sigma_j) < \dfrac{d-1}{m_0+d-1}+\dfrac{d(d-1)}{(m_0+d)(m_0+d-1)}$
It remains to be shown that the lower bound is achieved at $n\to \infty$.
Finally, again by @Willy's point, let's rewrite $t_k$ as
$t_k=\sigma_k\sigma_{k+1}\left(\dfrac{d-1}{d}\right)^{n-k-2}(m_{n-1}+d-1)=\frac{d(\sigma_k-\sigma_{k+1})}{m_{k+1}}\left(\dfrac{d-1}{d}\right)^{n-k-2}(m_{n-1}+d-1)$
$=\frac{d(\sigma_k-\sigma_{k+1})}{m_{0}\left(\dfrac{d}{d-1}\right)^{k+1}}\left(\dfrac{d-1}{d}\right)^{n-k-2}(m_{n-1}+d-1)$
$=\frac{d(\sigma_k-\sigma_{k+1})}{m_{0}}\left(\dfrac{d-1}{d}\right)^{n-1}(m_{n-1}+d-1)$
$\implies \sum\limits_{k=1}^{n-2}t_k=\dfrac{d}{m_0}\left(\dfrac{d-1}{d}\right)^{n-1}(m_{n-1}+d-1)\sum\limits_{k=1}^{n-2}(\sigma_k-\sigma_{k+1})$
$=\dfrac{d}{m_0}\left(\dfrac{d-1}{d}\right)^{n-1}\left(m_{0}\left(\dfrac{d}{d-1}\right)^{n-1}+d-1\right)(\sigma_1-\sigma_{n-1})$
$=\left(d+\dfrac{d(d-1)}{m_{0}}\left(\dfrac{d}{d-1}\right)^{n-1}\right)(\sigma_1-\sigma_{n-1})$
$\therefore \lim\limits_{n\to \infty}\sum\limits_{k=1}^{n-2} \sigma_k\prod\limits_{j=k+1}^{n-1}(1-\sigma_j) =  \lim\limits_{n\to\infty}\sum\limits_{k=1}^{n-2}t_k$
$=\lim\limits_{n\to\infty}\left(d+\dfrac{d(d-1)}{m_{0}}\left(\dfrac{d}{d-1}\right)^{n-1}\right)\times\lim\limits_{n\to\infty}(\sigma_1-\sigma_{n-1})=d\sigma_1=\dfrac{d-1}{m_0+d-1}$