I have considered taking, $x_{n}=(-1)^{n}/n$, then by the alternating series test it can be shown that :
$$ \sum_{n=1}^{\infty} x_{n} \quad \text{is convergent}$$
while the product
$$\prod_{n=1}^{\infty}\bigg(1+\frac{(-1)^{n}}{n}\bigg) \quad \text{diverges to zero} $$ where $p_{n}=(1+(-1)/1)\cdots (1+(-1)^{n}/n)=0 \implies p_{n}\to 0$ therefore divergent to zero
is the example correct ?
This has appeared here before, but it is not easy to search.
Here is an example. $$ x_n = -1+\exp((-1)^n/\sqrt{n}) \\ x_n = -1 + \left(1 + \frac{(-1)^n}{\sqrt{n}}+\frac{1}{2n}+O(n^{-3/2})\right) \\ \sum_{n=1}^N x_n = \sum_{n=1}^N \frac{(-1)^n}{\sqrt{n}} +\frac{1}{2}\sum_{n=1}^N \frac{1}{n} + \sum_{n=1}^N O(n^{-3/2}) \\ \sum_{n=1}^\infty x_n\quad\text{diverges to } +\infty. $$ On the other hand $$ 1+x_n = \exp((-1)^n/n^{1/2}) \\ \log(1+x_n) = \frac{(-1)^n}{n^{1/2}} \\ \sum\log(1+x_n)\quad\text{ converges} \\ \prod(1+x_n) = \exp \sum\log(1+x_n)\quad\text{ converges} $$