Leibniz Alternating Series Test and Conditional Convergence

The setup

An alternating series has the form

$\sum_{n=1}^{\infty} (-1)^{n} b_n \quad\text{or}\quad \sum_{n=1}^{\infty} (-1)^{n-1} b_n,\qquad b_n > 0,$

so consecutive terms flip sign. The Leibniz (alternating series) test gives a remarkably cheap convergence criterion.

Theorem (Leibniz test)

The series $\sum (-1)^{n} b_n$ converges provided all three hold from some index $N$ onward:

1. Positive: $b_n > 0$,
2. Monotone decreasing: $b_{n+1} \le b_n$,
3. Vanishing: $\displaystyle\lim_{n\to\infty} b_n = 0$.

If condition 3 fails ($b_n \not\to 0$), the term test already forces divergence. Conditions 1-2 only need to hold eventually (for large $n$); the first finitely many terms never affect convergence.

A bonus: the partial sums bracket the true sum, giving the error bound $\left|S - S_N\right| \le b_{N+1}$ — the first omitted term.

Absolute vs. conditional convergence

Convergence by Leibniz does not mean the series converges absolutely. Classify in two passes:

So the recipe is: first check $\sum |a_n| = \sum b_n$. If it converges, you are done (absolute). If it diverges, fall back to the Leibniz test on the signed series to decide between conditional and divergent.

How to apply (checklist)

1. Expose the sign. Rewrite the numerator so the alternation is explicit. A common konkur trick: $\sin\!\big((n+\tfrac12)\pi\big) = \cos(n\pi) = (-1)^n$, and $\cos(n\pi) = (-1)^n$.
2. Isolate $b_n$ (the positive magnitude) and verify $b_n \downarrow 0$ for Leibniz.
3. Test $\sum b_n$ for absolute convergence (p-test, comparison, integral test).
4. Combine via the table above.

Worked mini-example

Classify $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}$.

• $b_n = \dfrac{1}{\sqrt{n}}$ is positive, decreasing, and $\to 0$, so by Leibniz the series converges.
• Absolute values: $\sum \dfrac{1}{\sqrt{n}} = \sum n^{-1/2}$ is a $p$-series with $p=\tfrac12 \le 1$, so it diverges.

Converges but not absolutely $\Rightarrow$ conditionally convergent.

The same logic settles $\sum \dfrac{(-1)^n}{\ln(\ln n)}$: $b_n=1/\ln(\ln n)\downarrow 0$ (Leibniz converges), yet $\ln(\ln n)$ grows so slowly that $\sum 1/\ln(\ln n)$ diverges by comparison with the harmonic series — hence again conditional.

Exam trick: Leibniz only certifies convergence of the signed series; it says nothing about absolute convergence. Whenever a question asks for the exact type, you must run a second test on $\sum |a_n|$. Skipping that step is how the 'absolutely convergent' distractor steals points from a series that is merely conditional.