Convergence question Suppose that $\{a_n\}$ is decreasing sequence of real numbers with $\lim_{n\to \infty} a_n=0$
Show that $$\Pi _{n=1}^{\infty}[1+(-1)^na_n]$$ converges $\iff$ $$\sum_{a=1}^{\infty}a_n^2$$ converges. 
 A: If the product is considered convergent if the sequence of partial products converges to $0$, the assertion is wrong.
Consider then $a_1 = 0$ and $a_{2n} = a_{2n+1} = \dfrac{1}{\sqrt{n+1}}$ for $n \geqslant 1$. The product then becomes
$$\lim_{N\to\infty}\prod_{k=2}^N \left(1+\frac{1}{\sqrt{k}}\right)\left(1 - \frac{1}{\sqrt{k}}\right) = \lim_{N\to\infty}\prod_{k=2}^N\left(1-\frac1k\right) = \lim_{N\to\infty} \frac1N = 0,$$
and $\sum a_n^2 = +\infty$.
So let's assume that the product is only considered convergent if the limit is nonzero (assuming no factor is $0$; one usually  allows finitely many $0$ factors and considers a product convergent if the product of the remaining factors converges to a nonzero limit).
Then we may without loss of generality assume $0 \leqslant a_n < 1$ for all $n$, and the sequence of partial products
$$P_N = \prod_{n=1}^N \left(1 + (-1)^na_n\right)$$
converges to a nonzero number. Since all $P_N$ are positive, that is the case if and only if the sequence of logarithms,
$$\log P_N = \sum_{n=1}^N \log \left(1 + (-1)^na_n\right)$$
converges.
A Taylor expansion of the logarithms and the alternating series criterion then yield the result.
A: Expanding on the first answer, and the idea to use the binomial formula: Assuming that $a_n<1$, then because of monotonicity, one can bracket the given product by the product with alternate signs,
$$\prod_{n=1}^\infty (1-(-1)^na_n)$$
Then the original product converges if and only if the product of both converges, which is
$$\prod_{n=1}^\infty (1-a_n^2)$$
where the convergence to a non-zero term now follows from the standard result using the generalized Bernoulli inequality.
