If $\sum_{n=1}^{\infty} a_n$ converges and $\sum_{n=1}^{\infty} a_n^2$ diverges, then $\prod_{n=1}^{\infty} (1+a_n)$ diverges to $0$.

Question

Prove that if $\sum_{n=1}^{\infty} a_n$ converges and $\sum_{n=1}^{\infty} a_n^2$ diverges, then $\prod_{n=1}^{\infty} (1+a_n)$ diverges to $0$.

We assume that $\{a_n\}$ is a real sequence.

An infinite product $\prod_{n=1}^{\infty} (1+a_n)$ diverges to $0$ if $1+a_n \ne 0 \,\forall n \in \mathbb{N}$ and $\prod_{n=1}^{\infty} (1+a_n)=0$.

Any hint or reference will be appreciated.

Answer 1

Hint: $\log(1+t) \sim t - t^2/2$ as $t \to 0$ so $t - t^2/4 \ge \log(1+t)$ for real $t$ with $|t|$ sufficiently small (in fact it is true for $-1 < t < 2.532354756$ approximately). So when $a_i, i = k \ldots m$ are in this interval

$$ \prod_{i=k}^m (1 + a_i) = \exp \left(\sum_{i=k}^m \log(1 + a_i) \right) \le \ldots $$