A Question on the Convergence of Products I may be asking a very basic question in real analysis, but I can neither prove or disprove it even after repeated trials. The question is:
Let $a_j$ be a sequence of non-negative real numbers such that $\prod_{j=1}^n (1+a_j)$ is bounded as a sequence of $n$. Prove or disprove that $\prod_{j=1}^n (1+2a_j)$ is also bounded as a sequence of $n$. 
Even if this is true, does there exist a large constant $M$, such that $\prod_{j=1}^n (1+Ma_j)$ is unbounded as a sequence of $n$?
 A: The answers are yes, it is true; and no, there is no such $M$.
First, by monotone convergence you can replace every "bounded" in your question by "convergent."
Second, as mentioned in a comment of mine above:

If $\prod_{j=1}^n (1+a_j)$ bounded does imply $\prod_{j=1}^n (1+2a_j)$ bounded, then $\prod_{j=1}^n (1+a_j)$ bounded will imply $\prod_{j=1}^n (1+M a_j)$ for any fixed constant $M>0$. This is just by induction: for any fixed $k\geq0$, you'll have $\prod_{j=1}^n (1+a_j)$ bounded implies $\prod_{j=1}^n (1+2^k a_j)$ bounded.

So it is sufficient to show that if $(A_n)_n$ with $A_n\stackrel{\rm def}{=}\prod_{j=1}^n (1+a_j)$ converges, then so does $(B_n)_n$ with $B_n\stackrel{\rm def}{=}\prod_{j=1}^n (1+2a_j)$.
Now,
$$
\ln A_n = \ln \prod_{j=1}^n (1+a_j) = \sum_{j=1}^n \ln(1+a_j) < \infty \tag{1}
$$
and since this series converges, one must have in particular $\lim_{n\to\infty}\ln(1+a_n)=0$. This implies 
$$
\lim_{n\to\infty} a_n = 0. \tag{2}
$$
This will be really helpful: indeed, we now know that $\ln(1+a_n)= a_n + o(a_n)$ when $n\to\infty$, and therefore by comparison with (1) that
$$
\sum_{j=1}^n a_j < \infty  \tag{3}
$$
This allows us to conclude, as
$$
\ln B_n = \ln \prod_{j=1}^n (1+2a_j) = \sum_{j=1}^n \ln(1+2a_j)
\leq  \sum_{j=1}^n 2a_j < \infty
$$
where we used that $\ln(1+x)\leq x$, and (3).
Therefore, $(\ln B_n)_n$ converges (by monotone convergence, as it is a bounded series with non-negative terms) when $n\to \infty$, and so does $(B_n)_n$.
