# Find the limit of the sum of product expression $P_n = \sum_{j = 1}^{n-1} \prod_{i = j+1}^n (1-a_i) a_j$

Let $$\{a_n\}$$ be a convergent sequence, assuming each $$1 \geq a_i \geq 0$$

consider the series of products

$$P_n = \sum_{j = 1}^{n-1} \prod_{i = j+1}^n (1-a_i) a_j$$

I would like to find $$\lim_{n \to \infty} P_n$$ or if not possible, find an upperbound to this sequence.

Attempt:

$$\sum_{j = 1}^{n-1} \prod_{i = j+1}^n (1-a_i) a_j \leq \sum_{j = 1}^{n-1} \prod_{i = 1}^n (1-a_i) a_j$$

Wikipedia says $$\prod_{i = 1}^n (1-a_i) \leq \exp\left(-\sum_{i = 1}^n a_i\right)$$

So

$$\sum_{j = 1}^{n-1} \prod_{i = j+1}^n (1-a_i) a_j \leq \exp(-\sum_{i = 1}^n a_i) \sum_{j = 1}^{n-1} a_j$$

Then the upperbound to this limit is $$\exp\left(-\lim_{i = 1}^n \sum_{i = 1}^n a_i\right) \cdot\lim_{n \to \infty} \sum_{j = 1}^{n-1} a_j.$$

Is this correct? An alternative approach uses the AM-GM inequality on the inner product term.

• Does the product apply to just $(1-a_i)$ or to $(1-a_i)a_j$? IE $\prod_{i=2}^3 \left[ ( 1 - a_i) a_1 \right] = (1-a_2)a_1(1-a_3)a_1$. Commented Mar 11, 2022 at 23:53
• @CalvinLin According to his attempt, I think it is $P_n = \sum_{j = 1}^{n-1} a_j\prod_{i = j+1}^n (1-a_i)$. According to Le p'tit bonhomme's answer, it is equal to $1-a_n-\prod_{i=1}^n(1-a_i)$. Commented Mar 12, 2022 at 1:11

One can easily prove by induction that: $$P_n=1-a_n-\prod_{i=1}^n(1-a_i).$$
In the case where the limit $$\lim_{i\to+\infty} a_i=\alpha>0$$, the product tends to $$0$$ and $$\lim_{n\to+\infty} P_n=1-\alpha.$$ (Indeed, one can find a geometric upperbound of the product.)
In the case where the limit $$\lim_{i\to+\infty} a_i=0$$, the sequence of the products $$\prod_{i=1}^n(1-a_i)$$ is non negative and non increasing, so it converges to some $$\beta\in[0,1]$$ and then $$\lim_{n\to+\infty} P_n=1-\beta.$$
--$$a_i=0$$ implies $$\beta=1$$.
--$$a_i=\frac 1{i+1}$$ implies $$\beta=0$$.