Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n}. $$
Explanation of Method
Since $\ln n$ grows slower than $n$ itself, $\frac1{\ln n}$ shrinks even more slowly than $\frac1n$, and comparison shows $\sum\frac1{\ln n}$ diverges (in fact "more strongly" than the harmonic series). But $\frac1{\ln n}$ is positive, decreasing, and tends to $0$, so Leibniz's test still applies, giving conditional convergence exactly as with the plain alternating harmonic series.
- Show $\sum\frac1{\ln n}$ diverges via limit comparison with $\sum\frac1n$.
- Verify the three Leibniz conditions for $b_n=\frac1{\ln n}$.
- Conclude conditional (not absolute) convergence.
Step 1 · Test for absolute convergence
By L'Hôpital's rule, $\displaystyle\lim_{n\to\infty}\frac{n}{\ln n}=\lim_{n\to\infty}\frac{1}{1/n}=\lim_{n\to\infty}n=\infty$. Since $\sum\frac1n$ diverges and the ratio $\frac{1/\ln n}{1/n}=\frac n{\ln n}\to\infty$, by the limit comparison test $\sum\frac1{\ln n}$ diverges too (it is "more divergent" than the harmonic series). No absolute convergence.
Step 2 · Verify Leibniz's conditions for $b_n=\frac1{\ln n}$
For $n\ge2$, $\ln n>0$ so $b_n>0$. The function $f(x)=\frac1{\ln x}$ has $f'(x)=-\dfrac1{x(\ln x)^2}<0$ for $x\ge2$, so $b_n$ is strictly decreasing. And $\lim_{n\to\infty}\frac1{\ln n}=0$.
Step 3 · Apply Leibniz's test
All three conditions hold, so by Leibniz's test the series converges. Combined with Step 1, the series converges conditionally.
Summary Table
| Test | Result |
|---|---|
| $\sum1/\ln n$ (absolute series) | Diverges ($n/\ln n\to\infty$) |
| Leibniz's test | All conditions satisfied |
| Conclusion | Conditionally convergent |
Additional Notes
- Since $\frac1{\ln n}>\frac1n$ for $n\ge3$, this series' underlying positive sequence decays even more slowly than the harmonic series — yet Leibniz's test doesn't care about the rate of decay at all, only that it is monotone and tends to $0$, so convergence still holds.
- Compare with Problem 95's $\sum\frac1{n(\ln n)^\alpha}$: there, the extra factor of $\frac1n$ makes the series converge outright (for $\alpha>1$) even without alternation, while here the alternation is essential since $\sum\frac1{\ln n}$ alone diverges dramatically.
- This is the "slowest-decaying" example of conditional convergence among Problems 103, 104, 107, and 108 — the sequence $1/\ln n$ decreases so gradually that partial sums of the alternating series converge only very slowly to their limit, even though convergence is guaranteed.
Result
The series converges conditionally but not absolutely: