9
$\begingroup$

Suppose every term of a countably infinite sequence $x_1,x_2,\dots$ is between $0$ and $1$, i.e. $0<x_i<1$ for every $i$. Does $\mathop {\lim }\limits_{n \to \infty } \prod\limits_{i = 1}^n {{x_i}} $ equal $0$ or there is counterexample that it might not be $0$? Thank you!

$\endgroup$
1
14
$\begingroup$

Let $(a_i)$ be a strictly increasing sequence of positive numbers which is bounded above. Such a sequence necessarily converges, call the limit $a$. Now set $x_i = \frac{a_i}{a_{i+1}}$. Then $0 < x_i < 1$ and

$$\prod_{i=1}^nx_i = x_1\dots x_n = \frac{a_1}{a_2}\dots\frac{a_n}{a_{n+1}} = \frac{a_1}{a_{n+1}}.$$

Therefore,

$$\prod_{i = 1}^{\infty}x_i = \lim_{n\to\infty}\prod_{i=1}^n x_i = \lim_{n\to\infty}\frac{a_1}{a_{n+1}} = \frac{a_1}{a}$$

which is non-zero as $a_1 \neq 0$. It follows from this construction that any $p \in (0, 1)$ arises as such an infinite product.

$\endgroup$
10
$\begingroup$

$$\frac{1}{2}=\frac{1}{2^{1/2}}\cdot \frac{1}{2^{1/4}}\cdot \frac{1}{2^{1/8}}\cdot \frac{1}{2^{1/16}}\cdot \frac{1}{2^{1/32}}\cdot \frac{1}{2^{1/64}} \cdots$$

$\endgroup$
7
$\begingroup$

Start with the decreasing sequence $$p_1 = 1 - \frac{1}{4}, \quad p_2 = 1 - \frac{3}{8}, \quad p_3 = 1 - \frac{7}{16}\quad, \quad ...\quad , \quad p_n = 1 - \frac{2^n-1}{2^{n+1}} $$ whose limit equals $\frac{1}{2}$. Then set $$x_1 = p_1, \quad x_2 = p_2/p_1, \quad x_3 = p_3/p_2,\quad \ldots $$ The partial product $\prod_{i=1}^n x_i$ equals $p_n$, and so the limit is $\frac{1}{2}$.

$\endgroup$
5
$\begingroup$

Take $x_i = 1-\frac{1}{(i+1)^2}$. Then, we have $$ \ln \prod_{i=1}^n x_i = \sum_{i=1}^n \ln\left(1-\frac{1}{(i+1)^2}\right) $$ and as $\ln\left(1-\frac{1}{(i+1)^2}\right) \sim_{i\to\infty} -\frac{1}{i^2}$, the series $\sum_{i=1}^n \ln\left(1-\frac{1}{(i+1)^2}\right)$ converges (by comparison) to some real number $\ell < 0$. But this means that $$ \prod_{i=1}^n x_i\xrightarrow[n\to\infty]{} e^\ell > 0. $$

$\endgroup$
2
  • $\begingroup$ ...and then I think the product goes to $1/2$. $\endgroup$ Jan 12 '16 at 3:24
  • $\begingroup$ Possible... TBH, since convergence was sufficient, I did not care too much about the exact value. (Also, adding computations and details could actually drown the point...) $\endgroup$
    – Clement C.
    Jan 12 '16 at 3:25
3
$\begingroup$

Good result to know: Suppose $a_n\in (0,1)$ for all $n.$ Then $\prod_{n=1}^\infty (1-a_n) > 0$ iff $\sum a_n < \infty.$ The proof is a nice exercise in taking logs and using $\log (1+u) = u + o(u).$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.