Determine, for each a > 0, whether the series sum from n = 2 to infinity of ln(1 + (-1)^n / n^a) converges absolutely, converges conditionally, or diverges.

Step-by-step solution

1. 1

   Do not test the alternating part alone. The terms look alternating, and Leibniz's test tempts you immediately. But $\ln(1+x)$ is not an odd function: $\ln(1+t)$ and $\ln(1-t)$ do not cancel, and the leftover bias is what actually decides this problem. So expand instead of testing.

2. 2

   Expand the logarithm. With $x_n=(-1)^n n^{-\alpha}\to 0$ and $\ln(1+x)=x-\tfrac{x^2}{2}+O(x^3)$,

   $\ln\!\left(1+\frac{(-1)^n}{n^{\alpha}}\right)=\underbrace{\frac{(-1)^n}{n^{\alpha}}}_{\text{alternating}}-\underbrace{\frac{1}{2n^{2\alpha}}}_{\text{always negative}}+O\!\left(\frac{1}{n^{3\alpha}}\right)$

   The second term has no sign change at all — that is the hidden bias.

3. 3

   Handle the alternating piece. $\sum (-1)^n n^{-\alpha}$ converges for every $\alpha>0$ by Leibniz (the terms decrease monotonically to zero), and converges absolutely exactly when $\alpha>1$. So this piece never causes divergence.

4. 4

   Handle the quadratic piece — this is the decider. $\sum \tfrac{1}{2}n^{-2\alpha}$ is a $p$-series with $p=2\alpha$; it converges iff $2\alpha>1$, i.e. $\alpha>\tfrac12$. When $\alpha\le\tfrac12$ this piece diverges to $-\infty$ while the alternating piece stays bounded, so the whole series diverges.

5. 5

   Assemble the three regimes. The $O(n^{-3\alpha})$ remainder is dominated by the $n^{-2\alpha}$ term, so it never changes the verdict:

   • $\alpha>1$: both pieces converge absolutely, since $\left|\ln\left(1+\tfrac{(-1)^n}{n^\alpha}\right)\right|\sim n^{-\alpha}$ and $\sum n^{-\alpha}$ converges.
   • $\tfrac12<\alpha\le 1$: the series converges, but $\left|\ln(\cdot)\right|\sim n^{-\alpha}$ with $\alpha\le 1$ makes $\sum|\cdot|$ diverge — conditional convergence.
   • $0<\alpha\le\tfrac12$: divergent.
6. 6

   State the result.

   $\sum_{n=2}^{\infty}\ln\!\left(1+\frac{(-1)^n}{n^{\alpha}}\right)=\begin{cases}\text{absolutely convergent}, & \alpha>1\\[2pt]\text{conditionally convergent}, & \tfrac12<\alpha\le 1\\[2pt]\text{divergent}, & 0<\alpha\le\tfrac12\end{cases}$

   The moral: for $\ln(1+a_n)$ with alternating $a_n$, always expand to second order before applying an alternating-series test.