TITLE: Range of the map $(a_n)\to \prod_{n=1}^\infty(1+a_n).$
QUESTION [6 upvotes]: 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
$$P((a_n))= \prod_{n=1}^\infty(1+a_n).$$

REPLY [2 votes]: 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.