Let $x_n \in (0,1)$, is it possible that $\prod_{n=1}^\infty x_n >0$ ? I think it isn't, because such small numbers multiplied together will become smaller and smaller, but I am not sure if there is a positive lower bound for the product. Thanks!

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    $\begingroup$ @Dominique: Why does strictly decreasing imply the product is eventually arbitrarily small? $\endgroup$ – Ethan Feb 26 '13 at 16:19
  • $\begingroup$ Let me remove my comment. I think I answered that too fast! $\endgroup$ – Dominique Feb 26 '13 at 16:24
  • $\begingroup$ this question math.stackexchange.com/questions/508865/… provides another example $\endgroup$ – Rodrigo Jan 16 '19 at 18:28

For example, set $$a_n = 1+\frac{1}{2^n}$$

and then $$b_n = \frac{a_n}{a_{n-1}} < 1,$$ so that $$\prod_{k=1}^n b_k = \frac{a_n}{2}.$$

Of course, it implies that your $x_n \to 1$. In case $x_n \not\to 1$ it means that there exists $\alpha \in (0,1)$ such that $x_n < \alpha$ infinitely many times, and $\prod_k^n x_k \leq \alpha^{\#\{k \leq n \mid x_k < \alpha\}} \to 0$.

Hope that helps ;-)

  • $\begingroup$ I don't see very well why you make a difference between $b_k$ and $x_n$, can you just not take $x_n = b_n$ to solve the question? $\endgroup$ – Rodrigo Jan 16 '19 at 18:41

Well, if $\prod_n x_n = \theta$, and $x_n, \theta >0$, then $\ln(\prod_n x_n ) = \sum_n \ln x_n = \ln \theta$. Then you can use your knowledge of summations to find an example.

Here is one: $\theta = e^{-\frac{1}{2}}$, and $x_n = e^{-\frac{1}{2^{n+2}}}$.

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    $\begingroup$ Very nice answer, though I think it might be better phrased using $\exp$ instead of $\ln$. i.e. by continuity of $\exp(x)$, $\sum_{i=0}^{\infty}x_n \rightarrow x \implies \prod_{i=0}^{\infty}e^{x_n} \rightarrow e^x$. $\endgroup$ – Tom Oldfield Feb 26 '13 at 16:28
  • $\begingroup$ @TomOldfield: I wanted to connect the product to a sum. Your suggestion is tidier, but I think the connection less immediate? $\endgroup$ – copper.hat Feb 26 '13 at 16:34
  • $\begingroup$ I guess it just depends on how you look at it! I think of exponential rules more readily than logarithmic ones, I suppose. It was just a minor thing, and either way it's a lovely answer! $\endgroup$ – Tom Oldfield Feb 26 '13 at 16:37

Take your favorite converging series with positive general term $$ \sum_{n\geq 1}y_n=S. $$ Then set $$ x_n=e^{-y_n}=\frac{1}{e^{y_n}}. $$ You have $$ \prod_{n\geq 1}x_n=e^{-S}=\frac{1}{e^S}. $$


For $n\in\Bbb Z^+$ let $y_n=\prod_{k=1}^nx_k$. Define $x_n$ recursively so that $$y_n=\frac14+\frac1{2^{n+1}}\;.$$

Then $x_1=y_1=\frac14+\frac14=\frac12$, and for $n>1$ we must have $$x_n=\frac{y_n}{y_{n-1}}=\frac{\frac14+\frac1{2^{n+1}}}{\frac14+\frac1{2^n}}=\frac{2^{n-1}+1}{2^{n-1}+2}<1\;.$$

That is, the sequence $$\left\langle\frac{2^{n-1}+1}{2^{n-1}+2}:n\in\Bbb Z^+\right\rangle$$

is a sequence of numbers in $(0,1)$ whose product is


I picked the target product $\frac14$ arbitrarily and chose to approach it in steps of $2^{-k}$; the same idea can clearly be used to construct by brute force an example with any desired product in $(0,1)$, approached by steps of any reasonable sequence of sizes.


It's quite simple to see actually. For any product $\prod_{i=1}^\infty n_i$ where $n_i>1$ that converges (such as $\prod_{i=1}^\infty \frac{1}{i^2}+1 \approx 3.676 $), you can take the reciprocal and then it is obvious that $n_i<1$ but $\prod_{i=1}^\infty{n_i}>0$.


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