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In my real analysis class I was asked this which got me stick:

Is there an example of a sequence of real positive numbers increasing to the limit 1 $ \{ a_n \}_{n=1}^{\infty} $ such that the partial products $ a_1 , a_1a_2,a_1a_2a_3,... $ converges to a positive limit?

I thought about it and thought it might be true because I coud not disprove it generally but I cannot come up with an example. Woud someone please be able to provide an example if any? Thanks to all helpers.

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    $\begingroup$ By working with $b_n=\log a_n$, your problem is equivalent to construct a increasing sequence $(b_n)$ of negative real numbers converging to $0$ such that the series $\sum b_n$ converges, which is in turn equivalent (via $c_n=-b_n$) to construct a convergent series $\sum c_n$ with $(c_n)$ decreasing.. $\endgroup$
    – Matemáticos Chibchas
    Apr 17, 2016 at 3:22
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    $\begingroup$ Should have put that into an answer, would be worth much more than just giving a random example without telling OP how to get to it. $\endgroup$
    – Nobody
    Apr 17, 2016 at 7:17
  • $\begingroup$ @MatemáticosChibchas I second Nobody's suggestion. I think you should turn your comment into an answer. $\endgroup$
    – Mark S.
    Apr 17, 2016 at 18:25
  • $\begingroup$ @Nobody For future reference, if you use "@", you can ping a user with a comment. Otherwise only the person who made the post you are commenting will be pinged. See the following for more details: meta.stackexchange.com/questions/43019/… $\endgroup$
    – Mark S.
    Apr 17, 2016 at 18:27
  • $\begingroup$ Indeed, @MatemáticosChibchas's comment is a much better answer than the answers. $\endgroup$
    – R.. GitHub STOP HELPING ICE
    Apr 17, 2016 at 19:22

5 Answers 5

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Take any positive decreasing sequence $x_1 , x_2, \dots$ that converges to a positive value $x$
Define $a_n = \frac{x_{n+1}}{x_n}$

Then $$a_1 a_2 \dots a_n = \frac{x_{n+1}}{x_1}\to \frac{x}{x_1}>0$$

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  • $\begingroup$ Shouldn't it be $x_1$ in the denominator? $\endgroup$
    – Frxstrem
    Apr 17, 2016 at 14:22
  • $\begingroup$ @Frxstrem That's right. Thanks $\endgroup$
    – Kitegi
    Apr 17, 2016 at 17:01
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One may choose $$ a_n=1-\frac1{4n^2} $$ we have $$ \lim_{n\to \infty}a_n=1 $$ and we have $$ \lim_{N\to \infty}\prod_{n=1}^Na_n=\lim_{N\to \infty}\prod_{n=1}^N\left(1-\frac1{4n^2}\right)=\frac2{\pi}>0 $$ where we have used $$ \lim_{N\to \infty}\prod_{n=1}^N\left(1-\frac{x^2}{n^2}\right)=\frac{\sin \pi x}{\pi x}. $$

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$$ \begin{align} \prod_{k=1}^n\left(1-\frac1{(k+1)^2}\right) &=\prod_{k=1}^n\frac{k(k+2)}{(k+1)^2}\\ &=\prod_{k=1}^n\frac{k}{k+1}\prod_{k=1}^n\frac{k+2}{k+1}\\ &=\frac1{n+1}\frac{n+2}2\\[3pt] &\to\frac12 \end{align} $$

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Another example is given by $a_n = 1/(1+2^{-2^n})$.

The partial products in this case take the form $a_1...a_k=1/(1+1/4+1/16+\ldots+1/4^{2^k-1})$, converging to $1/\sum_{k=0}^\infty(1/4)^k = 3/4$.

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Solve this recursively (think through it). Say you want the product to be at least 0.5. So try $a_1 = 0.9$. How big does $a_2$ have to be now to keep the product about $0.5$? Pick an $a_2$ that works say $(1+0.5/0.9)/2$. Now how big does $a_3$ have to be? And so forth.

This process actually gives you an inductively defined sequence that solves your problem, but I'm sure (because I saw the other answers) you can find one with an explicit form (maybe by bounding the one you produce by one with an explicit form).

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