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

Put

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

so that

$a_j{b}_j=\frac{(-1{)}^j\iota (j+3)}{\iota (j+1{)}^2}\mathrm{.}$

Then ${\sum }_j{a}_j$ is the alternating harmonic series, which converges (${\sum }_{j\ge 0}(-1{)}^j\mathrm{/}(j+1)$ converges conditionally, with sum strictly between $1\mathrm{/}2$ 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 $\iota (j+3)\mathrm{/}\iota (j+1{)}^2$ 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 $b_j\to 1\mathrm{\ne }0$. Abel's test is designed for exactly this shape: a convergent series multiplied by a monotone bounded factor.

• Boundedness alone would not do. With $a_j=(-1{)}^j\mathrm{/}\sqrt{j+1}$ convergent and $b_j=(-1{)}^j$ bounded but not monotone, $\sum a_j{b}_j=\sum 1\mathrm{/}\sqrt{j+1}$ 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 $2\mathrm{/}\iota (j+1)\to 0$, 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 $\iota (j+3)\mathrm{/}\iota (j+1{)}^2\ge 1\mathrm{/}\iota (j+1)$, so the series of absolute values dominates the harmonic series; nothing below needs this, and it is recorded to place the example alongside ${\sum }_{j\ge 0}(-1{)}^j\mathrm{/}(j+1)$ converges conditionally, with sum strictly between $1\mathrm{/}2$ and $1$ rather than alongside Every rearrangement of ${\sum }_{k\ge 0}(-1\mathrm{/}2{)}^k$ converges to $2\mathrm{/}3$.