Confirm a seemingly wrong series convergence result I read this on a paper: Given a sequence $\{p_n\}$, where $0<p_n<1$, if $\sum_{n=1}^{\infty}p_n<\infty$, $\sum_{n=1}^{\infty}\log(1-p_n)>-\infty$, i.e. the sum here does not go to negative infinity. 
I don't think this is true because for any $x \in (0,1)$, $-x>\log(1-x)$, which means $\sum_{n=1}^{\infty}-p_n>\sum_{n=1}^{\infty}\log(1-p_n).$ There might be some choice of $p_n$ to provide a counter example but I'm not certain. I don't see an obvious choice here.
Can anyone shed some light on this result?
 A: First of all, I don't see a contradiction between your statement and the statement of the theorem, because you can still have $\sum_{n=1}^\infty \log(1-p_n)>-\infty$ even if $\sum_{n=1}^\infty -p_n>\sum_{n=1}^\infty \log(1-p_n)$, these two statements are certainly not mutually exclusive.
To prove the statement, you can write $\sum_{n=1}^\infty \log(1-p_n)=\log\Big(\prod_{n=1}^\infty (1-p_n)\Big)=-\log\big(\prod_{n=1}^\infty (1-p_n)^{-1}\big)$.  Now we have 
\begin{equation}
\log\big(\prod_{n=1}^\infty (1-p_n)^{-1}\big)\leq \log\big(e^{\sum_{n=1}^\infty ((1-p_n)^{-1}-1)}\big)=\sum_{n=1}^\infty \Big((1-p_n)^{-1}-1\Big)=\sum_{n=1}^\infty\sum_{m=1}^\infty p_n^m.
\end{equation}
Now consider the series $\sum_{m=1}^\infty\sum_{n=1}^\infty p_n^m$, if this converges than so does the last series above. Define the sum $c_m=\sum_{n=1}^\infty p_n^m$ so that the sum in question can be written $\sum_{m=1}^\infty c_m$.  Define the largest element of $p_n$ as $p^*$. It is clear that $\sum_{n=1}^\infty p_n^{m+1}\leq p^*\sum_{n=1}^\infty p_n^{m}$, so $\frac{c_{m+1}}{c_m}<p^*<1$.  Therefore, by the ratio test the series converges, so we have 
\begin{equation}
-\log\big(\prod_{n=1}^\infty (1-p_n)^{-1}\big)\geq \sum_{n=1}^\infty\sum_{m=1}^\infty p_n^m>-\infty,
\end{equation}
which implies the result.
