Request help proving the following:

(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?

  • 1
    $\begingroup$ 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). $\endgroup$ – Clement C. Mar 16 '18 at 21:26
  • 1
    $\begingroup$ Your alternative request is answered by the limit comparison test, with some caveats like Clement's comment above. $\endgroup$ – Fimpellizieri Mar 16 '18 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 $\sum_{n=k_0}^{\infty}\ln(1-a_n)$

  • $\begingroup$ Good job! ${}$ +1 $\endgroup$ – Fimpellizieri Mar 16 '18 at 21:42
  • $\begingroup$ @Mostafa Ayaz very nice, thanks $\endgroup$ – user2661923 Mar 16 '18 at 21:48
  • $\begingroup$ @Clement C. very interesting, thanks $\endgroup$ – user2661923 Mar 16 '18 at 21:49
  • $\begingroup$ @Fimpellizieri that was just what I wanted, thanks. $\endgroup$ – user2661923 Mar 16 '18 at 21:54
  • $\begingroup$ ??? 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. $\endgroup$ – zhw. Mar 16 '18 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.

  • $\begingroup$ creative and slick. if i hadn't already "accepted" 1st answer, i would have accepted yours. $\endgroup$ – user2661923 Mar 16 '18 at 23:59

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