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.

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.
