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I'm thinking about a question:

Suppose $y_n > −1$ for all $n$ and $\sum |y_n| < \infty$. Show that $\prod_{m=1}^\infty (1 + y_m)$ exists.

Since $\sum |y_n| < \infty$, we must be able to find a constant $M = \sup(1 + y_n)$ with $M < \infty$. Let $X_0 = 1$ and $X_n = \prod_{m=1}^n (1 + y_m) /M^n$. $X_n$ forms a super martingale. It follows from martingale convergence theorem that $X_n \to c$ almost surely, where $c$ is a constant with $0 \le c \le \mathbb E[X_0] = 1$. Since the limit of $M^n c$ exists, we must have the limit of $\prod_{m=1}^\infty (1 + y_m)$ also exists.

Is this a correct proof?


OK, I think this might not have much to do with martingale. Now second try.

It suffices to show that $\sum_{m=1}^\infty |\log(1+y_m)|$ converges. By Taylor's theorem, we have $$ \log(1+y_m) = \frac 1 {a_m} y_m, $$ where $a_m$ is a number between $1$ and $1 + y_m$. Since $\sum_m |y_m| < \infty$, we have $y_m \to 0$. So for any $L > 1$, we can find $M$ such that for all $m \ge M$, $|1/a_m| \le L$. Therefore, we have $$ \sum_{m \ge M} |log(1+y_m)| \le M \sum_{m \ge M} |y_m| < \infty. $$ Since dropping finite terms at the beginning of a series does not change weather it converges, we have $\sum_{m=1}^\infty |\log(1+y_m)|$ converges.

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I like it, a nifty idea, +1. –  gt6989b Feb 21 '13 at 0:19
    
Why does the limit of $M^n$ exist? –  Byron Schmuland Feb 21 '13 at 0:24
    
@ablmf But if $M^n\to\infty$, then you cannot conclude that $\prod_{m=1}^\infty (1 + y_m)$ exists. –  Byron Schmuland Feb 21 '13 at 0:32
    
@ByronSchmuland Yeah, you are right. There is a problem if $M > 1$. –  ablmf Feb 21 '13 at 0:34
    
@ablmf There is no harm in trying. Not every idea pans out, however! –  Byron Schmuland Feb 21 '13 at 0:37

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