Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(n+1) a^{2n}}. $$
Explanation of Method
Since the series involves $a^{2n}=(a^2)^n$ in the denominator, the ratio test on the absolute series naturally produces a limit of $1/a^2$, splitting the analysis into $|a|>1$, $|a|=1$, and $|a|<1$. At $|a|=1$ the series reduces exactly to the alternating harmonic-type series of Problem 103 with $\alpha=1$ (shifted by one index), while $|a|<1$ makes the terms blow up, forcing outright divergence.
- Apply the ratio test to the absolute series for general $|a|$.
- Resolve the boundary case $|a|=1$ using Leibniz's test.
- Resolve the case $|a|<1$ using the divergence (n-th term) test.
Step 1 · Ratio test on the absolute series (general $|a|$)
With $|a_n|=\dfrac1{(n+1)|a|^{2n}}$:
$$ \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} = \lim_{n\to\infty}\frac{n+1}{(n+2)|a|^2} = \frac1{|a|^2}. $$
If $|a|>1$: $\frac1{|a|^2}<1$, so the series converges absolutely. If $|a|<1$: $\frac1{|a|^2}>1$, so $\sum|a_n|$ diverges (ratio test).
Step 2 · Resolve $|a|=1$ using Leibniz's test
At $|a|=1$, $a^{2n}=1$, so the series becomes $\sum\frac{(-1)^{n-1}}{n+1}$. With $b_n=\frac1{n+1}$: $b_n>0$, decreasing, and $b_n\to0$. By Leibniz's test the series converges; since $\sum\frac1{n+1}$ (harmonic-type) diverges, this is conditional convergence.
Step 3 · Resolve $|a|<1$ using the divergence test
For $|a|<1$, $|a|^{2n}\to0$, so $|a_n|=\dfrac1{(n+1)|a|^{2n}}\to\infty$. Since $a_n\not\to0$, the series diverges by the $n$-th term test.
Summary Table
| Condition | Conclusion |
|---|---|
| $|a|>1$ | Converges absolutely |
| $|a|=1$ | Converges conditionally |
| $|a|<1$ | Diverges |
| $a=0$ | Undefined |
Additional Notes
- The threshold $|a|=1$ here plays the same structural role as the parameter thresholds in Problems 58–66's ratio-test series — a boundary case where the ratio test gives exactly $1$, requiring separate analysis (here, reduction to Leibniz's test rather than Raabe's test).
- This problem elegantly unifies three different convergence tests within one parametrized family: the ratio test (for $|a|\ne1$), Leibniz's test (at $|a|=1$), and the divergence test (for $|a|<1$) — each governing a different regime of the same underlying series.
- The case $|a|=1$ specializing to $\sum\frac{(-1)^{n-1}}{n+1}$ is, up to a one-index shift, exactly Problem 103's series with $\alpha=1$: both illustrate the same conditional-convergence phenomenon at the harmonic-series threshold.
Result
Convergence depends entirely on $|a|$: