# 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?

• Your alternative request is false in general, and to be true the statement should require that at least one of the two sequences $(a_n)_n$, $(b_n)_n$ stays positive for $n$ big enough (or stays negative; but basically does not change sign). Commented Mar 16, 2018 at 21:26
• Your alternative request is answered by the limit comparison test, with some caveats like Clement's comment above. Commented Mar 16, 2018 at 21:28

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 $\sum_{n=k_0}^{\infty}\ln(1-a_n)$

• Good job! ${}$ +1 Commented Mar 16, 2018 at 21:42
• @Mostafa Ayaz very nice, thanks Commented Mar 16, 2018 at 21:48
• @Clement C. very interesting, thanks Commented Mar 16, 2018 at 21:49
• @Fimpellizieri that was just what I wanted, thanks. Commented Mar 16, 2018 at 21:54
• ??? You started with an inequality that says that a negative number is less than a positive number, and ended with an infinite series of negative numbers being less than an infinite series of positive numbers.
– zhw.
Commented Mar 16, 2018 at 23:06

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.

• creative and slick. if i hadn't already "accepted" 1st answer, i would have accepted yours. Commented Mar 16, 2018 at 23:59