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Let $a_n$ be real $|a_n|<1$, $\sum a_n$ converges. Then show that $ \prod (1+a_n)$ converges if and only if $\sum a_n^2$ converges.

What I know about infinite product is that for complex $a_n \neq -1$, $\prod(1+a_n)$ converges if and only if $\sum Log(1+a_n)$ (Log: principal branch) converges.(and also some about absolute convergence) But I don't know how to use this.

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Let $\sum_na_n^2$ converge. We have

$$\log(1+a_n)=a_n-\frac12a_n^2+\frac13\frac1{(1+\xi_n)^3}a_n^3\;,$$

where $\xi_n$ lies between $0$ and $a_n$. Thus we can split $\sum_n\log(1+a_n)$ into three series. The first two converge by hypothesis. In the third, since $\sum_na_n$ converges, $\xi_n$ goes to $0$, so the term containing it is bounded by a constant. Also $\sum_n|a_n^3|$ is dominated by $\sum_na_n^2$, so the third series converges absolutely and thus converges; hence $\sum_n\log(1+a_n)$, being the sum of three convergent series, also converges.

Conversely, let $\sum_na_n^2$ diverge. Since $\sum_na_n$ converges, $a_n$ goes to $0$, so for sufficiently large $n$

$$ -\frac12a_n^2+\frac13\frac1{(1+\xi_n)^3}a_n^3\lt-\frac14a_n^2\;. $$

Thus the sum over these two terms diverges, and since $\sum_na_n$ converges, it follows that $\sum_n\log(1+a_n)$, being the sum of a convergent series and a divergent series, diverges.

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