Abel's Test

Put $A_n = \sum_{k=1}^{n} a_k$.

By hypothesis $A_n \to A$.

Set $A_0 := 0$ so the formula below is also valid for $m = 1$ (with $A_{m-1} = A_0 = 0$).

Summation by parts (Abel's transform) gives, for $m \le n$, $\sum_{k=m}^{n} a_k b_k = A_n b_{n+1} - A_{m-1} b_m + \sum_{k=m}^{n} A_k (b_k - b_{k+1})$, which is checked directly by substituting $a_k = A_k - A_{k-1}$ and collecting the telescoping terms.

Write $A_k = A + \varepsilon_k$ with $\varepsilon_k \to 0$. Then $\sum_{k=m}^{n} A_k (b_k - b_{k+1})$ splits as $A(b_m - b_{n+1}) + \sum_{k=m}^{n} \varepsilon_k (b_k - b_{k+1})$, the first part telescoping.

The key is the monotonicity of $(b_k)$: the differences have constant sign, so $\sum_{k=m}^{n} |b_k - b_{k+1}| = |b_m - b_{n+1}|$.

Boundedness $|b_k| \le M$ makes this at most $2M$.

Hence $\left|\sum_{k=m}^{n} \varepsilon_k (b_k - b_{k+1})\right| \le \max_{m \le k \le n} |\varepsilon_k| \cdot 2M$.

For any $\eta > 0$, taking $m$ large makes this smaller than $\eta \cdot 2M$.

Now the boundary term $A_n b_{n+1} - A_{m-1} b_m$.

Since $b_k \to b$ (a monotone bounded sequence converges), this tends to $A b - A b = 0$ in the Cauchy sense as $m, n \to \infty$.

The leftover term $A(b_m - b_{n+1})$ must also be estimated: $|A(b_m - b_{n+1})| \le |A| \cdot |b_m - b_{n+1}|$, and since $b_k \to b$, $|b_m - b_{n+1}| \to 0$ as $m, n \to \infty$ (crudely $|b_m - b_{n+1}| \le 2M$).

Hence this term also tends to $0$.

Every term in the decomposition is now estimated, so together with the previous estimate, the tails of $\sum_{k=m}^{n} a_k b_k$ become arbitrarily small, so the partial sums of $\sum a_k b_k$ form a Cauchy sequence.

Hence the series converges.

Set $a_k = (-1)^k/k$.

Then $\sum a_k$ converges by the alternating series (Leibniz) test, with sum $-\ln 2$.

And $b_k = 2 + 1/k$ is decreasing and satisfies $2 \le b_k \le 3$, hence bounded.

The hypotheses of Abel's test are satisfied verbatim, so $\sum_{k=1}^{\infty} a_k b_k = \sum_{k=1}^{\infty} \frac{(-1)^k}{k}\left(2 + \frac{1}{k}\right)$ converges.

Why Dirichlet's test does not apply directly: its hypothesis demands $b_k \to 0$ monotonically, but here $b_k \to 2 \ne 0$. One may of course split the series as $2\sum_{k=1}^{\infty} \frac{(-1)^k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^k}{k^2}$ and prove convergence of each, but that is a separate workaround argument — not a direct application of Dirichlet's test.

The exact value is available too: $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^2} = -\eta(2) = -\frac{\pi^2}{12}$, so the limit is $-2\ln 2 - \pi^2/12 \approx -2.2088$ — matching the mini-visual, where $S_{500} \approx -2.2068$.

Set $a_k = (-1)^k/\sqrt{k}$.

Since $1/\sqrt{k} \downarrow 0$, $\sum a_k$ converges by the Leibniz test (conditionally).

For $b_k = \frac{k}{k+1}$, the inequality $b_k < b_{k+1}$ cross-multiplies to $k(k+2) < (k+1)^2$, i.e. $k^2 + 2k < k^2 + 2k + 1$, i.e. $0 < 1$ — so $(b_k)$ is increasing.

Since $b_k = \frac{k}{k+1} = 1 - \frac{1}{k+1}$, we have $0 < b_k < 1$ — bounded, with $\lim b_k = 1 \ne 0$. The "$b_k \to 0$" hypothesis of Dirichlet's test fails again.

Since $\sum a_k$ converges and $(b_k)$ is monotone and bounded, Abel's test gives convergence of $\sum_{k=1}^{\infty} \frac{(-1)^k}{\sqrt{k}} \cdot \frac{k}{k+1}$. Numerically: $S_{120} \approx -0.2090$, $S_{400} \approx -0.2292$, $S_{500} \approx -0.2318$ — the tail settles.