$\prod_{j \ge 0} \bigl(1 + (-1)^{j}/\sqrt{j+2}\bigr)$ has partial products tending to $0$ although $\sum_{j \ge 0} (-1)^{j}/\sqrt{j+2}$ converges

For every $j$, $\iota(j+2) \ge \iota(2) = 2$, so $\sqrt{\iota(j+2)} \ge \sqrt 2 > 1$ and $|p_j| = 1/\sqrt{\iota(j+2)} \le 1/\sqrt 2 < 1$; hence every factor satisfies $0 < f_j \le 1 + 1/\sqrt 2 < 2$, and every $\Pi_m$ is positive.

Fix $i$ and put $a := \iota(2i+2)$, $b := \iota(2i+3)$, so $b - a = 1$ and $a + b = \iota(4i+5)$, both positive. By [L1], $f_{2i} = 1 + 1/\sqrt a$ and $f_{2i+1} = 1 - 1/\sqrt b$.

An induction gives that finite products are monotone in nonnegative factors: if $0 \le x_i \le y_i$ for all $i < n$ then $\prod_{i<n}x_i \le \prod_{i<n}y_i$, since both products are nonnegative and $\prod_{i<n+1}x_i = \bigl(\prod_{i<n}x_i\bigr)x_n \le \bigl(\prod_{i<n}y_i\bigr)x_n \le \bigl(\prod_{i<n}y_i\bigr)y_n$.

The sequence $\beta_j := 1/\sqrt{\iota(j+2)}$ is positive, nonincreasing and converges to $0$: monotonicity from $0 < \iota(j+2) < \iota(j+3)$ and strict increase of the square root, and convergence because, given a rational $\varepsilon > 0$, an $n \ge 1$ with $1/\iota(n) < \varepsilon^{2}$ gives $\iota(j+2) > (1/\varepsilon)^{2}$ and so $\beta_j < \varepsilon$ for every $j \ge n$.

Since $\sqrt b - \sqrt a = (b-a)/(\sqrt a + \sqrt b) = 1/(\sqrt a + \sqrt b)$, one has $\dfrac1{\sqrt a} - \dfrac1{\sqrt b} = \dfrac{\sqrt b - \sqrt a}{\sqrt{a}\sqrt{b}} = \dfrac{1}{\sqrt{ab}\,(\sqrt a + \sqrt b)}$, so $P_i := f_{2i}f_{2i+1} = 1 + \dfrac1{\sqrt a} - \dfrac1{\sqrt b} - \dfrac1{\sqrt{ab}} = 1 - \dfrac{1}{\sqrt{ab}}\Bigl(1 - \dfrac{1}{\sqrt a + \sqrt b}\Bigr)$.

Here $\sqrt a \ge \sqrt 2 > 1$ and $\sqrt b \ge \sqrt 3 > 1$, so $\sqrt a + \sqrt b > 2$ and $1 - 1/(\sqrt a + \sqrt b) > 1/2$; and by [L5], $\sqrt{ab} \le (a+b)/2 = \iota(4i+5)/2$, so $1/\sqrt{ab} \ge 2/\iota(4i+5)$.

An induction gives $\Pi_{2n} = \prod_{i<n} P_i$ for every $n$: at $n = 0$ both are the empty product $1$, and $\Pi_{2(n+1)} = \Pi_{2n} f_{2n} f_{2n+1} = \Pi_{2n}P_n$.

By the alternating series test $\sum_j p_j = \sum_j \varepsilon_j \beta_j$ converges.

Combining, $P_i \le 1 - \bigl(2/\iota(4i+5)\bigr)\cdot\tfrac12 = 1 - 1/\iota(4i+5) \le 1 - q_i$, where $q_i := 1/\iota(4i+6)$, using $\iota(4i+5) < \iota(4i+6)$; and $0 < P_i$ by step 1.1, while $0 < q_i < 1$.

Hence $0 < \Pi_{2n} = \prod_{i<n}P_i \le \prod_{i<n}(1 - q_i)$ for every $n$.

The series $\sum_i q_i$ diverges: $6(i+1) = 6i+6 \ge 4i+6$, so $\iota(4i+6) \le 6\,\iota(i+1)$ and $q_i \ge \tfrac16\cdot\dfrac1{\iota(i+1)}$; the series $\sum_i \tfrac16 \cdot 1/\iota(i+1)$ diverges, being a nonzero multiple of the harmonic series, so $\sum_i q_i$ diverges by comparison.

By [L8] applied to $(q_i)$, the partial products $\prod_{i<n}(1-q_i)$ tend to $0$; with step 4.1 and the squeeze, $\Pi_{2n} \to 0$.

Also $\Pi_{2n+1} = \Pi_{2n} f_{2n}$ with $0 < f_{2n} < 2$, so $0 < \Pi_{2n+1} < 2\,\Pi_{2n}$ and $\Pi_{2n+1} \to 0$ as well.

Therefore $\Pi_m \to 0$: given a rational $\varepsilon > 0$, choose $N$ with $\Pi_{2n} < \varepsilon/2$ for all $n \ge N$; then for $m \ge 2N+1$, writing $m$ as $2n$ or $2n+1$ according to the partition of $\mathbb{N}$ by the two index maps, in either case $n \ge N$ and $\Pi_m \le 2\Pi_{2n} < \varepsilon$.

For every $N'$ the $N'$-th tail products satisfy $\prod_{j=N'}^{N'+n-1}f_j = \Pi_{N'+n}/\Pi_{N'}$ with $\Pi_{N'} > 0$ fixed, so they tend to $0$ too; no tail has partial products with a nonzero limit, and $\prod_j (1+p_j)$ diverges.

So $\sum_j p_j$ converges while $\prod_j (1+p_j)$ diverges, and the refuted claim fails; the hypothesis it is missing is a sign condition, or absolute convergence of $\sum p_j$, as in For $p_k \ge 0$ the product $\prod (1 + p_k)$ converges iff $\sum p_k$ converges, with $1 + \sum_{k<n} p_k \le \prod_{k<n}(1+p_k) \le 1/\bigl(1 - \sum_{k<n} p_k\bigr)$ when $\sum_{k<n} p_k < 1$; for $0 \le p_k < 1$ the product $\prod (1 - p_k)$ converges iff $\sum p_k$ converges and its partial products tend to $0$ otherwise; and $\sum |p_k|$ convergent implies $\prod (1+p_k)$ convergent.