Computing a limit of a sum mixed with product.

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.

Solution 1:

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

$t_k=\sigma_k\sigma_{k+1}\left(\dfrac{d-1}{d}\right)^{n-k-2}(m_{n-1}+d-1)=\frac{d(\sigma_k-\sigma_{k+1})}{m_{k+1}}\left(\dfrac{d-1}{d}\right)^{n-k-2}(m_{n-1}+d-1)$

$=\frac{d(\sigma_k-\sigma_{k+1})}{m_{0}\left(\dfrac{d}{d-1}\right)^{k+1}}\left(\dfrac{d-1}{d}\right)^{n-k-2}(m_{n-1}+d-1)$

$=\frac{d(\sigma_k-\sigma_{k+1})}{m_{0}}\left(\dfrac{d-1}{d}\right)^{n-1}(m_{n-1}+d-1)$

$\implies \sum\limits_{k=1}^{n-2}t_k=\dfrac{d}{m_0}\left(\dfrac{d-1}{d}\right)^{n-1}(m_{n-1}+d-1)\sum\limits_{k=1}^{n-2}(\sigma_k-\sigma_{k+1})$

$=\dfrac{d}{m_0}\left(\dfrac{d-1}{d}\right)^{n-1}\left(m_{0}\left(\dfrac{d}{d-1}\right)^{n-1}+d-1\right)(\sigma_1-\sigma_{n-1})$

$=\left(d+\dfrac{d(d-1)}{m_{0}}\left(\dfrac{d}{d-1}\right)^{n-1}\right)(\sigma_1-\sigma_{n-1})$

$\therefore \lim\limits_{n\to \infty}\sum\limits_{k=1}^{n-2} \sigma_k\prod\limits_{j=k+1}^{n-1}(1-\sigma_j) = \lim\limits_{n\to\infty}\sum\limits_{k=1}^{n-2}t_k$

$=\lim\limits_{n\to\infty}\left(d+\dfrac{d(d-1)}{m_{0}}\left(\dfrac{d}{d-1}\right)^{n-1}\right)\times\lim\limits_{n\to\infty}(\sigma_1-\sigma_{n-1})=d\sigma_1=\dfrac{d-1}{m_0+d-1}$

Solution 2:

Let $a=\frac d{d-1}$ and we need to compute $$P_n=\prod_{k=1}^n \left(1-\frac{1}{m_0\, a^k+d}\right)$$ Using Pochhamer symbols $$P_n=\frac {d+m_0}{d+m_0-1}\left(\frac{d-1}{d}\right)^{n+1}\frac{\left(-\frac{m_0}{d-1};a\right)_{n+1}}{\left(-\frac{m_0}{d};a\right)_{n+1}}$$ Replace $a$ by its value and simplify to obtain $$P_n=\frac {1}{m_0+d-1}\left(\frac{d-1}{d}\right)^{n}\left(m_0 \left(\frac{d}{d-1}\right)^n+d-1\right)=\frac {m_0+(d-1)\left(1-\frac{1}{d}\right)^n}{m_0+d-1}$$ Therefore $$P_\infty=\frac{m_0}{m_0+d-1}$$

Checking for $m=10$ and $d=3$, the $P_n$ make the sequence $$\left\{\frac{17}{18},\frac{49}{54},\frac{143}{162},\frac{421}{486},\frac{1247}{1458}, \frac{3709}{4374},\frac{11063}{13122},\frac{33061}{39366},\frac{98927}{118098},\frac{296269}{354294},\frac{887783}{1062882},\frac{2661301}{3188646}\right\}$$ The last value $$P_{12}=\frac{2661301}{3188646}=\frac 56+\frac{2048}{1594323}=\frac 56+0.00129$$ $$P_{24}=\frac{1412164459621}{1694577218886}=\frac 56+\frac{8388608}{847288609443}=\frac 56+0.00001$$

I cannot do the second one.