Prove that $\prod_{n \in \mathbb{N}}{(1-a_n)} \geq 1 - \sum_{n \in \mathbb{N}}{a_n}$

Question

The following proof is obtained from this paper.

My question is how to obtain the inequality.

My guess is because of the following inequality:

$$\prod_{n \in \mathbb{N}}{(1-a_n)} \geq 1 - \sum_{n \in \mathbb{N}}{a_n}$$

By the condition given in the statement, we know that $\sum_{n \in \mathbb{N}}{a_n}$ converges. Thus, the product $\prod_{n \in \mathbb{N}}{(1-a_n)}$ converges. But I don't know how to prove the inequality.

Any hint would be much appreciated.

Answer 1

Induction on $n$. If $0\leq a_j\leq 1$ for $j\in N$ then we have $\prod_{j=1}^n(1-a_j)\geq 1-\sum_{j=1}^na_j$ for every $n\in N.$ PROOF. The case $n=1$ is trivial.If case $n$ is true, then $$\prod_{j=1}^{n+1}(1-a_j)=[\;\prod_{j=1}^n(1-a_j)\;]\cdot(1-a_{n+1})\geq$$ $$\geq [1-\sum_{j=1}^na_j\;]\cdot(1-a_{n+1})=[1-\sum_{j=1}^na_j]-a_{n+1}+a_{n+1}\sum_{j=1}^na_j\geq$$ $$\geq [1-\sum_{j=1}^na_j]-a_{n+1}=1-\sum_{j=1}^{n+1}a_j$$ so case $n+1$ is true.

Answer 2

Suppose $\prod_{k=1}^n (1-a_k) \ge 1- \sum_{k=1}^n a_k$, and $a_{n+1} \in (0,1)$, then $(1-a_{n+1}) \prod_{k=1}^n (1-a_k) \ge (1-a_{n+1})(1- \sum_{k=1}^n a_k) = 1-\sum_{k=1}^{n+1} a_k + a_{n+1} \sum_{k=1}^n a_k \ge 1- \sum_{k=1}^{n+1} a_k$.

Answer 3

First, note the following. $$ \Pi_{n\in \mathbb{N}} (1-a_n) = \lim_{k\rightarrow \infty}\Pi_{i=0}^{k} (1-a_n) $$ and similarly $$ 1-\sum_{n\in \mathbb{N}} a_n = 1-\lim_{k\rightarrow \infty}\sum_{i=0}^{k} a_n. $$ Let $$ b_k = \lim_{k\rightarrow \infty}\Pi_{i=0}^{k} (1-a_n) $$ and $$ c_k = 1-\lim_{k\rightarrow \infty}\sum_{i=0}^{k} a_n. $$ The claim that you are asking follows if we can show that $b_k\geq c_k$ for every $k\in \mathbb{N}$. This can be proven by induction. We would need to look at a sequence of statements of the following sort. $$ P(k) : b_k\geq c_k \text{ for all }k. $$ Let's take $P(0)$ as the base case. This is easy to see because $b_0 = c_0 = 1-a_0$, confirming $P(0)$.

Next, we will try to prove that $P(k)$ implies $P(k+1)$. Note that $b_{k+1} = b_k \times (1-a_{k+1})$ and $c_{k+1} = c_k-a_{k+1}$. Consider the following. $$ c_{k+1} = c_k-a_{k+1} = c_k - a_{k+1}\frac{c_k}{c_k} = c_k\left( 1 - \frac{a_{k+1}}{c_k} \right) $$ For simplicity let's assume that $c_k>0$, because otherwise the sequence of $c_k$ would be less than 0, which automatically satisfies the original claim. Nowe notice that $c_k<1$, hence $$ \left( 1 - \frac{a_{k+1}}{c_k} \right)<1-a_{k+1}. $$ As a result, since $b_k>c_k$ by assumption, we have $$ b_{k+1}= b_k \times (1-a_{k+1}) > c_k \times\left( 1 - \frac{a_{k+1}}{c_k}\right)= c_{k+1} $$ which proves the conjecture.