This question is from Ponnusamy and silvermann complex variables with applicatio Subsection ( Infinite Products ) and I am struck in 1 part of it.
Suppose that ${a_n}$ is a decreasing sequence of real numbers with lim $ n\to \infty$ $ a_n =0$ . Show that $$\prod_{n=1}^{\infty}[1+ (-1)^n a_n] $$ converges iff $$\sum_{n=1}^{\infty} {a_n}^2 $$ converges.
I have done the part which assumes convergence of $$\sum_{n=1}^{\infty} {a_n}^2 .$$
For converse , convergence of $$\prod_{n=1}^{\infty}[1+ (-1)^n a_n]$$ imples convergence of series $$\sum_{n=1}^{\infty} (-1)^n a_n$$ converges (let to a). Now , I am unable to use convergence of $$\sum_{n=1}^{\infty} (-1)^n a_n$$ to a to prove convergence of $$\sum_{n=1}^{\infty} {a_n}^2 .$$
Kindly help me with that
We have $a_n \searrow 0$ and, hence, $$\lim_{n \to \infty}\frac{(-1)^na_n - \log(1+(-1)^n a_n)}{a_n^2} = \frac{1}{2}$$
This implies that for all sufficiently large $n \geqslant N$,
$$\frac{(-1)^n a_n - \log(1+(-1)^na_n)}{a_n^2} > \frac{1}{4},$$
and
$$\sum_{n=N}^M(-1)^n a_n - \sum_{n=N}^M\log(1+(-1)^na_n) > \frac{1}{4}\sum_{n-N}^Ma_n^2$$
Note that $\sum (-1)^n a_n$ converges by the alternating series test. If $\sum a_n^2$ diverges, then it diverges to $+\infty$ since each term is nonnegative. This would imply
$$\log \prod_{n=N}^\infty (1+ (-1)^na_n) =\sum_{n=N}^\infty\log(1+(-1)^na_n) = -\infty,$$
and the infinite product diverges to zero, a contradiction.
The convention for infinite products is that $\prod_{n=1}^\infty (1+a_n) = 0$ is considered divergence.