$\sum_{j \ge 0} (-1)^{j}\,(j+3)/(j+1)^{2}$ converges, by Abel's test with the monotone bounded factor $(j+3)/(j+1)$

Put

$a_j := \frac{(-1)^j}{\iota(j+1)}, \qquad b_j := \frac{\iota(j+3)}{\iota(j+1)} \qquad (j \in \mathbb{N}),$

so that

$a_j b_j \;=\; \frac{(-1)^j\,\iota(j+3)}{\iota(j+1)^{2}} .$

Then $\sum_j a_j$ is the alternating harmonic series, which converges ($\sum_{j \ge 0} (-1)^{j}/(j+1)$ converges conditionally, with sum strictly between $1/2$ and $1$), while $(b_j)$ is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and bounded, with $1 < b_j \le 3$. By Abel's test (Abel's test: if $\sum a_k$ converges and $(b_k)$ is monotone and bounded then $\sum a_k b_k$ converges) the series

$\sum_{j \ge 0} \frac{(-1)^{j}\,(j+3)}{(j+1)^{2}}$

converges.

Neither of the two earlier tests reaches it as directly. The alternating series test: if $(b_k)$ is nonincreasing with $b_k \to 0$ then $\sum_{k} (-1)^{k} b_k$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_n$ would require the sequence $\iota(j+3)/\iota(j+1)^{2}$ to be nonincreasing, which is true but is an extra computation; Dirichlet's test: if the partial sums of $\sum a_k$ are bounded and $(b_k)$ is nonincreasing with $b_k \to 0$, then $\sum a_k b_k$ converges would require a factor tending to $0$, and $b_j \to 1 \ne 0$. Abel's test is designed for exactly this shape: a convergent series multiplied by a monotone bounded factor.

• Boundedness alone would not do. With $a_j = (-1)^{j}/\sqrt{j+1}$ convergent and $b_j = (-1)^{j}$ bounded but not monotone, $\sum a_j b_j = \sum 1/\sqrt{j+1}$ diverges takes a convergent series and a bounded factor that is not monotone, and the product series diverges. Here monotonicity of $(b_j)$ is what step 2.1 supplies, and it is the hypothesis doing the work.

• The limit of the factor is $1$, not $0$, and that is the point. Since $2/\iota(j+1) \to 0$, the factor tends to $1$, so Dirichlet's test: if the partial sums of $\sum a_k$ are bounded and $(b_k)$ is nonincreasing with $b_k \to 0$, then $\sum a_k b_k$ converges does not apply to $(b_j)$ directly. Abel's test converts this into Dirichlet's by writing $b_j = (b_j - 1) + 1$, which is exactly what its proof does.

• The series converges conditionally. Its terms in absolute value are $\iota(j+3)/\iota(j+1)^{2} \ge 1/\iota(j+1)$, so the series of absolute values dominates the harmonic series; nothing below needs this, and it is recorded to place the example alongside $\sum_{j \ge 0} (-1)^{j}/(j+1)$ converges conditionally, with sum strictly between $1/2$ and $1$ rather than alongside Every rearrangement of $\sum_{k \ge 0} (-1/2)^{k}$ converges to $2/3$.