User @Math Lover provided an example where $ \sum_{n=0}^{\infty} q_n = \infty$ but $ \sum_{n=0}^{N} \log(1+q_n)$ alternates between $\log 2$ and $0$ in $N$. In this answer, we construct an example where
$$ \sum_{n=0}^{\infty} q_n = \infty \quad \text{and} \quad \sum_{n=0}^{\infty} \log(1+q_n) = 0.$$
To this end, we prepare two auxiliary sequences:
- $(\epsilon_k)_{k=1}^{\infty}$ is a sequence such that $1 \geq \epsilon_k \geq \epsilon_{k+1} > 0$ and $\epsilon_k \to 0$ as $k \to \infty$.
- $(N_k)_{k=1}^{\infty}$ is a sequence of positive integers such that $ N_k \epsilon_k^2 \geq 1$ for all $k$.
Now define $(r_n)$ and $(q_n)$ by
$$ (r_n)_{n=0}^{\infty} = ( \underbrace{ \epsilon_1, -\epsilon_1, \cdots, \epsilon_1, -\epsilon_1 }_{2N_1\text{-terms}}, \underbrace{ \epsilon_2, -\epsilon_2, \cdots, \epsilon_2, -\epsilon_2 }_{2N_2\text{-terms}}, \cdots ),
\qquad q_n = e^{r_n} - 1. $$
Then $r_n = \log(1+q_n)$, and
By the alternating series test, $\sum_{n=0}^{\infty} r_n$ converges. Moreover, its odd-th partial sums are identically zero, hence the sum is also zero.
Using the fact that $e^x \geq 1 + x + \frac{1}{e}x^2$ for $|x| \leq 1$, it follows that
$$ q_n \geq r_n + \frac{1}{e}r_n^2 $$
Now summing over $n$, we obtain
$$ \sum_{n=0}^{\infty} q_n \geq \left( \sum_{n=0}^{\infty} r_n \right) + \left( \sum_{k=1}^{\infty} 2N_k \cdot \frac{\epsilon_k^2}{e} \right) \geq \sum_{k=1}^{\infty} \frac{2}{e} = \infty. $$