# Range of the map $(a_n)\to \prod_{n=1}^\infty(1+a_n).$

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

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

• 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.
– user751478
Commented Feb 20, 2020 at 18:55
• I'm curious: Why did this get a downvote?
– zhw.
Commented Feb 20, 2020 at 23:13
• 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. Commented Feb 20, 2020 at 23:53
• 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 $A$?
– BCLC
Commented Dec 1, 2020 at 0:08

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

• "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?
– zhw.
Commented Feb 20, 2020 at 23:12
• @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. Commented Feb 20, 2020 at 23:49
• @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. Commented Feb 20, 2020 at 23:52
• @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.
– zhw.
Commented Feb 21, 2020 at 22:53
• I gave an answer myself, which takes a different tack on some parts of the problem.
– zhw.
Commented Feb 23, 2020 at 18:56

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 $$P((a_n)) for all $$(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.

• 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
– BCLC
Commented Dec 1, 2020 at 0:10