Let $y_1, y_2, ....$ be a sequence such that $0\leq y_n \leq 1$ and $\sum_{n=1}^\infty y_n=\infty$. Prove that $\prod_{n=1}^\infty (1-y_n)=0.$ Let $y_1, y_2, ....$ be a sequence such that $0\leq y_n \leq 1$ for all $n$, and
$\displaystyle\sum_{n=1}^{\infty} y_n=\infty$.  Prove that $\displaystyle\prod_{n=1}^{\infty} (1-y_n)=0.$

Then, "\begin{eqnarray}
 \ln \left[\displaystyle\prod_{n=s}^{t} f(n)\right] &=& \displaystyle\sum_{n=s}^{t}\left[\ln f(n)\right], \\
1-x &\leq & e^{-x}. 
\end{eqnarray}
\begin{eqnarray*}
 \displaystyle\prod_{n=1}^{\infty} (1-y_n) &=& {\Large e}^{\left[\ln \displaystyle\prod_{n=1}^{\infty} (1-y_n)\right]} \\
   &\leq& {\Large e}^{\left[\ln \left(\displaystyle\prod_{n=1}^{\infty} e^{-y_n}\right)\right]}  \\
   &=& {\Large e}^{\left(\displaystyle\sum_{n=1}{-y_n}\right)}\\
    &=& {\Large e}^{\left(-\displaystyle\sum_{n=1}{y_n}\right)} \ \left( \displaystyle\sum_{n=1}^{\infty} y_n=\infty \right)\\
   &=& {\Large e}^{(- \infty)} \\
   &=& 0.
\end{eqnarray*}"
Okay?
And thanks for watching the accounts.
 A: A more refined proof is as follows: let  $p_n = \displaystyle \prod_{k=1}^n (1-y_k), s_n = \displaystyle \sum_{k=1}^n y_k$, and use $1-y_k \le e^{-y_k}\implies 0\le p_n \le e^{-\displaystyle \sum_{k=1}^n y_k}= \dfrac{1}{e^{s_n}}$. Since $s_n \to +\infty \implies e^{s_n} \to +\infty\implies \dfrac{1}{e^{s_n}} \to 0\implies p_n \to 0$ which is the desire claim.
A: We will prove this by using the Borel-Cantelli theorem.
Let $X_i$ be independent random variables such that $\mathbb{P}(X_i=1)=y_i$ and 
$\mathbb{P}(X_i=0)=1-y_i$.
If $\sum_{k=1}^{\infty}y_k=\infty$, then by Borel-Cantelli $\{X_i=1~\text{i.o.}\}$ almost surely. Therefore $\mathbb{P}( X_i=0~ \forall i)=0$ and from independence we have $\mathbb{P}( X_i=0~ \forall i)= \prod_k (1-y_k)$ hence we conclude.
A: If any $y_n = 1$ then clearly the result is true, so suppose $y_n \in [0,1)$. We have $\log(1+x) \le x$ hence
$\log(1-y_n) \le -y_n$ and hence $\sum_n \log(1-y_n) = \log(\prod_n (1-y_n)) \to
 -\infty$ from
which the result follows.
A: I don't see any reason to introduce exponential or logarithm functions. Assume that $0\le x<1$. Then $1/(1-x) \ge (1+x)$. Assume $0\le y<1$. Then $(1+x)(1+y)\ge 1+(x+y)$.
Now if $0\leq y_n < 1$ for all $n$, we get that $$\prod_{n=1}^\infty \frac{1}{1-y_n} \ge 1+\sum_{n=1}^\infty y_n$$ and since we are given that the summation has infinite sum,  then so does the product of reciprocals. Hence, the original infinite product is zero.
Notice the requirement for $y_n<1$. If any of the $y_n=1$ then, according to some definitions, the infinite product is not defined.
