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
$$P((a_n))= \prod_{n=1}^\infty(1+a_n).$$
 A: 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, $P(A) = [2,e)$
A: 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.
