Prove that ${\prod_\limits{n = k_0}^{\infty} (1 - a_n)} \;$ converges to positive value Request help proving the following:
Given: 
(1) For all $n > 0 , \;a_n > 0 $ 
(2) ${\sum_\limits{n=1}^{\infty}a_n} \;$ is convergent. 
(3) There exists $k_0 \in \mathbb{Z^+} \; $ such that for all 
$ k \geq k_0 , \; a_k < 1.$ 
To prove: $\displaystyle{\prod_{n = k_0}^{\infty} (1 - a_n)} \;$ converges
to a $\lambda$ that is strictly greater than $0$.
Motive: On pages 10-11 of this pdf, the assertion helps prove that if 
${\sum_\limits{n=1}^{\infty}a_n} \;$ is convergent, then the infinite
continued fraction given by $[a_0; a_1, a_2, \cdots]$ does not converge to a
value.
Partial Work: My query has actually already been answered by 
this question.  Reviewing the answers in the previous question, I agree 
that $\ln(1-a_n) \sim -a_n \ (n \rightarrow +\infty).$  I also agree that it is 
sufficient to show that $\sum_\limits{k\geq k_0} \ln(1 - a_k)$ is convergent.
Stumbling Block: I googled on "real analysis equivalence test convergence"
and could not find any online reference.  I am actually requesting an 
$\epsilon,\delta$ argument that 
${\sum_\limits{k\geq k_0}a_k} \;$ convergent implies 
${\sum_\limits{k\geq k_0}\ln(1 - a_k)} \;$ convergent.
Alternative Request: An online reference to the equivalence test or an 
$\epsilon,\delta$ argument that : 
given two series $\{a_k\}$ and $\{b_k\}$ where 
$\lim_\limits{n\rightarrow\infty} \frac{a_n}{b_n} = c \neq 0,$ then $\sum a_k$ convergent iff $\sum b_k$ convergent? 
 A: By Taylor series we know that$$\ln(1-a_n)=a_n+\dfrac{a_n^2}{2}+\dfrac{a_n^3}{3}+...$$also we know that $a_n\to0$ therefore $$\exists b\in\Bbb N\qquad,\qquad \forall n>b\to 0<a_n<\dfrac{1}{2}$$which leads to $$0<a_n^2<\dfrac{a_n}{2}\\0<a_n^3<\dfrac{a_n^2}{2}<\dfrac{a_n}{4}\\0<a_n^4<\dfrac{a_n^3}{2}<\dfrac{a_n}{8}\\.\\.\\.$$which leads to$$\ln(1-a_n)=a_n+\dfrac{a_n^2}{2}+\dfrac{a_n^3}{3}+...<a_n+\dfrac{a_n}{2\times 2}+\dfrac{a_n}{2^2\times 3}+\dfrac{a_n}{2^3\times 4}+\dfrac{a_n}{2^4\times 5}+...<2a_n$$for $n>b$. Therefore $$\sum_{n=b+1}^{\infty}\ln(1-a_n)<\sum_{n=b+1}^{\infty}2a_n$$which concludes that $\sum_{n=b+1}^{\infty}\ln(1-a_n)$ is convergent and so is $\sum_{n=k_0}^{\infty}\ln(1-a_n)$
A: Because $\ln '(1)=1,$ we have $\ln(1+u)/u\to 1$ as $u\to 0.$ Thus there exists $\delta >0$ such that
$$\tag 1 \frac{\ln (1+u)}{u}< 2\,\, \text { for }|u|<\delta.$$
Now $\sum a_n <\infty$ implies $a_n\to 0.$ Hence there exists $N_0$ such that $n\ge N_0$ implies $0<a_n<\delta.$ For such $n$ we then have by $(1)$ that
$$\ln(1-a_n) > -2a_n\,\implies\,
\sum_{n=N_0}^{\infty} \ln (1-a_n) \ge \sum_{n=N_0}^{\infty} -2a_n.$$
The series on the right converges. Hence the so does the series on the left, being a sum of negative terms. This implies the desired result.
