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Let $A$ be the set of nonnegative sequences $(a_n)$ such that $\sum_{n=1}^{\infty}a_n=1.$ Find the range of the map $P:A\to \mathbb R$ defined by

$$P((a_n))= \prod_{n=1}^\infty(1+a_n).$$

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    $\begingroup$ If we take $0=a_{n+1}=a_{n+2}=...$, then $(1+a_1)(1+a_2)...(1+a_n)\leq\left(\frac{(1+a_1)+...+(1+a_n)}{n}\right)^n=\left(1+\frac{1}{n}\right)^n\leq e$. Also $\prod(1+a_n)\geq 1+\sum_n a_n=2$. These suggest trying to prove that $[2,e)$ are the values that we can get. Perhaps $e$ too. $\endgroup$
    – user751478
    Commented Feb 20, 2020 at 18:55
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    $\begingroup$ I'm curious: Why did this get a downvote? $\endgroup$
    – zhw.
    Commented Feb 20, 2020 at 23:13
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    $\begingroup$ If I were to take a guess, it's because the problem is posed without any context or attempt on your part. I didn't even know it did have a downvote, though, since it's at +5 right now. $\endgroup$ Commented Feb 20, 2020 at 23:53
  • $\begingroup$ borel-cantelli? seems familiar to a question i've asked before. as it turns out... $a_n \in [0,1]$ for all $a_n$ for all sequences in $A$? $\endgroup$
    – BCLC
    Commented Dec 1, 2020 at 0:08

2 Answers 2

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For this first part, let $(a_n) \in A$ be arbitrary.

  • By AM-GM and basic approximations for $e$, we have $$\prod_{n=1}^N(1+a_n) \leq \left(\frac{\sum_{n=1}^N(1+a_n)}{N}\right)^N = \left(1 + \frac{\sum_{n=1}^Na_n}{N}\right)^N \leq \left(1 + \frac{1}{N}\right)^N \leq e$$ Then by taking the limit as $N\to \infty,$ we have $P((a_n)) \leq e.$

  • Since $a_n \geq 0$ for all $n,$ it follows that $$P((a_n)) = \prod_{n=1}^\infty(1+a_n) \geq 1 + \sum_{n=1}^\infty a_n = 2$$ so we have $P((a_n)) \geq 2.$

From this, we know that $P(A) \subseteq [2,e].$


  • For each $k,$ if $a^k_1=a^k_2=\cdots=a^k_k = \frac{1}{k}$ and $a^k_i = 0$ for $i>k$, then $$P((a^1_n)) = 2 \\ \lim_{k\to\infty}P((a^k_n)) = \lim_{k\to\infty} \left(1+\frac{1}{k}\right)^k = e$$
  • Given any two sequences $(a_n), (b_n) \in A$ and $0 \leq t \leq 1,$ we have $((1-t)a_n+tb_n) \in A,$ and $$t \mapsto P(((1-t)a_n+tb_n))$$ is a continuous function, so the range of $P$ is an interval.

From this, we know $[2,e) \subseteq P(A).$


To finish, towards a contradiction suppose $P(A) = [2,e]$ so that $P((a_n)) = e$ for some $(a_n) \in A.$ Let $i$ be some index such that $a_i \neq a_1$ and define a new sequence by $$b_n = \begin{cases}\dfrac{a_1+a_i}{2} & \text{ if } n \in \{1,i\} \\ a_n & \text{ otherwise}\end{cases}$$ Then $(b_n) \in A$ and $$P((b_n)) = P((a_n)) \cdot \frac{\left(1 + \frac{a_1+a_i}{2}\right)^2}{(1+a_1)(1+a_i)} > P((a_n)) = e$$ which contradicts $P(A) \subseteq [2,e].$


That is, $P(A) = [2,e)$

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  • $\begingroup$ "Given any two sequences $(a_n), (b_n) \in A$ and $0 \leq t \leq 1,$ we have $((1-t)a_n+tb_n) \in A,$ and $$t \mapsto P(((1-t)a_n+tb_n))$$ is a continuous function," Why is that a continuous function? $\endgroup$
    – zhw.
    Commented Feb 20, 2020 at 23:12
  • $\begingroup$ @zhw. I'll try to write out some details for that. Note that we don't need the full strength of that statement, though, as it's a polynomial function (hence continuous) when $(a_n), (b_n) \in \{(a^k_n) : k \geq 1\}$ for the sequences $(a^k_n)$ defined in the previous step. That's all that is actually needed for the conclusion of that section. $\endgroup$ Commented Feb 20, 2020 at 23:49
  • $\begingroup$ @zhw I think you can even show that function is absolutely continuous, showing that the quantity $P((1-t)a_n+tb_n)\sum_{n=1}^{\infty}\frac{b_n-a_n}{(1-t)a_n+tb_n}$ is bounded above and below, and then integrating term by term to show the integral wrt $t$ is the function itself. $\endgroup$ Commented Feb 20, 2020 at 23:52
  • $\begingroup$ @BrianMoehring Yes, the not full strength version would be enough. I think it would be good to mention this in your solution. Which is a nice solution btw, +1. And a √ if you address the continuity issue. $\endgroup$
    – zhw.
    Commented Feb 21, 2020 at 22:53
  • $\begingroup$ I gave an answer myself, which takes a different tack on some parts of the problem. $\endgroup$
    – zhw.
    Commented Feb 23, 2020 at 18:56
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I thought I would give an answer too. First, some inequalites: If $x\ge 0,$ then

$$\tag 1 \ln (1+ x) \le x,$$

with strict inequality if $x>0.$ If $x,y\ge 0,$ then

$$\tag 2 |\ln (1+ y)-\ln (1+x)| \le |y-x|.$$

Both $(1),(2)$ follow readily from the MVT.

Now let $(a_n)\in A.$ Then $(1)$ gives

$$\ln P((a_n)) = \ln \left (\prod_{n=1}^\infty (1+a_n)\right) = \sum_{n=1}^{\infty}\ln (1+a_n) < \sum_{n=1}^{\infty}a_n=1.$$

Therefore $P((a_n))<e$ for all $(a_n)\in A.$

Next, note $P(1, 0, 0, 0, 0, \dots)=2$ and $P((a_n))\ge 2$ for all $(a_n)\in A$ as Brian Moehring showed.

Finally, to show the range of $P$ is the full interval $[2,e),$ I did this: We can view $A$ as a subset of $l^1(\mathbb N).$ In fact it's a convex subset of that normed linear space. Hence $A$ is connected in that space. Thus if we can show $P$ is continuous on $A$ with respect to the $l^1(\mathbb N)$ metric, then $P(A) = [2,e).$

To do this, it's enough to show $\ln P$ is continuous on $A.$ So suppose $(a_n),(b_n)\in A.$ Then

$$|\ln P((b_n))-\ln P((a_n))| = |\sum_{n=1}^{\infty}(\ln (1+b_n) - \ln (1+a_n))|$$ $$ \le \sum_{n=1}^{\infty}|\ln (1+b_n) - \ln (1+a_n)| \le \sum_{n=1}^{\infty}|b_n-a_n|.$$

We used $(2)$ to get to the last sum, which by definition is $\|(b_n)-(a_n)\|_{l^1(\mathbb N)}.$ Thus $\ln P$ is continuous on $A$ (actually Lipschitz), finishing the proof.

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  • $\begingroup$ i swear this is getting more and more familiar in re probability. i swear i've asked a familiar, but maybe not really related, question $\endgroup$
    – BCLC
    Commented Dec 1, 2020 at 0:10

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