A proposition about two sequences let's consider a sequence $0<a_1\leq a_2\leq....$ which is bounded by some constant M, i.e. $a_i\leq M$, $\forall i\in \mathbb{N}$. Now I want to prove the following:
The sequence $D_n:=a_1 \dotsb a_n$ converges to a nonzero limit if and ony if $\sum_{n=1}^\infty (1-a_n)<\infty$.
Unfortunately I couldn't find a way to prove this. Do you have some idea how to attack this problem?
Regards
 A: We can prove a more general result here.
At first we define $\prod_{k=1}^\infty a_k$ to be absolute convergent iff $\sum_{k=1}^\infty \log(a_k)$ is absolute convergent. That this definition makes sense you can see in this way:
If $s=\prod_{k=1}^\infty a_k$ is the limit and not equal to 0 we already know that $a_k$ converges to $1$ as we see that
$$\lim_{k\to \infty} a_k = \lim_{k\to \infty}\frac{\prod_{j=1}^{k} a_j}{\prod_{j=1}^{k-1} a_j}=\frac{s}{s}=1$$
Furthermore we will show that $\prod_{k=1}^\infty a_k$ converges to a non zero limit iff 
$\sum_{k=1}^\infty \log(a_k)$ converges. Assume $\sum \log(a_k)$ converges. Then we see using
the definitions $s_n = \sum_{k=1}^n \log (a_k)$ and $p_n = \prod_{k=1}^n a_k$, that $\exp(s_n)=p_n$
and because of the continuity of the exponential function 
$$\exp( \lim s_n)=\lim \exp(s_n)=\lim p_n.$$ 
The other direction is nearly exactly the same, even if you $a_k$ is a sequence of complex numbers.
In our situation we see that $a_n$ converges to $1$ monotone from below hence $\sum \log(a_n)$ converges
iff $\sum |\log(a_n)|$ converges.
Assuming $|z|<1$ we see that 
\begin{align*}
\left| 1 - \frac{\log(1+z)}{z}\right| &= | \tfrac{z}{2}-\tfrac{z^2}{3}+\dots |\\
&\leq \tfrac{1}{2} (|z|+|z|^2+\dots)\\
&= \frac{1}{2} \cdot \frac{|z|}{1-|z|}
\end{align*}
holds. Further if we say that $|z|< \frac{1}{2}$ even
$$ \left| 1- \frac{\log(1+z)}{z} \right| \leq \frac{1}{2}$$
holds and with the triangle inequality
$$ |a|-|b| \leq |a-b| \leq |a|+|b|$$
we get 
\begin{equation}
 \tfrac{1}{2}\cdot  |z|\leq |\log(1+z)| \leq \tfrac{3}{2}\cdot  |z|.
\end{equation}
The rest is just using the dominating convergence theorem. 
