Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \ln\!\left(1+\frac{\sin\frac{n\pi}4}{n^\alpha}\right), \qquad \alpha>0. $$
Explanation of Method
Since $\sin\frac{n\pi}4/n^\alpha\to0$ for $\alpha>0$, the Taylor expansion $\ln(1+x)=x-\frac{x^2}2+O(x^3)$ splits the term into a leading Dirichlet-type piece $\sin\frac{n\pi}4/n^\alpha$ (convergent for every $\alpha>0$, by Problem 132's logic) plus a correction $-\frac12\sin^2\frac{n\pi}4/n^{2\alpha}+O(n^{-3\alpha})$ whose convergence hinges on $2\alpha>1$, i.e. $\alpha>\frac12$. Below that threshold the correction term diverges and dominates, dragging the whole series to divergence despite the leading term's convergence.
- Expand $\ln(1+x)$ to second order in $x=\sin\frac{n\pi}4/n^\alpha$.
- Classify the leading term via Dirichlet's test (converges for all $\alpha>0$).
- Classify the quadratic correction term via comparison with $\sum n^{-2\alpha}$, and combine.
Step 1 · Taylor expansion
With $u_n=\sin\frac{n\pi}4/n^\alpha\to0$: $$ \ln(1+u_n) = u_n - \frac{u_n^2}2 + O(u_n^3) = \frac{\sin\frac{n\pi}4}{n^\alpha} - \frac{\sin^2\frac{n\pi}4}{2n^{2\alpha}} + O\!\left(\frac1{n^{3\alpha}}\right). $$
Step 2 · Leading term via Dirichlet's test
By Problem 132's argument (bounded partial sums of $\sin\frac{n\pi}4$, monotone $n^{-\alpha}\downarrow0$), $\sum\sin\frac{n\pi}4/n^\alpha$ converges for every $\alpha>0$ (conditionally for $0<\alpha\le1$, absolutely for $\alpha>1$).
Step 3 · Correction term and combination
$\sin^2\frac{n\pi}4\ge0$ and is not identically zero (period-$8$ pattern, e.g. $\sin^2(\pi/4)=\frac12$), so $\sum\sin^2\frac{n\pi}4/(2n^{2\alpha})$ compares directly (positive terms) with $\sum n^{-2\alpha}$: convergent iff $2\alpha>1$, i.e. $\alpha>\frac12$. The $O(n^{-3\alpha})$ remainder converges absolutely whenever $3\alpha>1$, always true once $\alpha>\frac12$. So: for $\alpha>1$, both leading and correction terms converge absolutely — absolute convergence. For $\frac12<\alpha\le1$, the leading term converges conditionally and the correction converges absolutely, so the sum converges but not absolutely (since $|\ln(1+u_n)|\sim|u_n|$ for small $u_n$, matching the leading term's absolute divergence) — conditional. For $0<\alpha\le\frac12$, the correction term diverges (to $+\infty$, all terms nonnegative) while the leading term converges, so the overall series diverges.
Summary Table
| Range of $\alpha$ | Conclusion |
|---|---|
| $\alpha>1$ | Absolutely convergent |
| $\frac12<\alpha\le1$ | Conditionally convergent |
| $0<\alpha\le\frac12$ | Diverges |
Additional Notes
- This is a direct sequel to Problems 132 and 133: the logarithm's nonlinearity introduces the new threshold $\alpha=\frac12$ that has no counterpart in the pure sine/cosine series, coming entirely from the quadratic term in the Taylor expansion — a pattern echoed in Problem 135's square-root perturbation problem.
- The technique of splitting a perturbed alternating-type term into a leading Dirichlet-convergent piece plus a sign-definite correction series is the standard approach whenever a nonlinear function (log, square root, power) is applied to a small oscillating quantity; the correction series' exponent is typically double (or otherwise scaled from) the original.
- Unlike Problems 132/133 where the conditional-convergence range extended down to $\alpha=0^+$, here the range is truncated at $\alpha=\frac12$ — the logarithm's concavity effectively "costs" convergence at low $\alpha$ that the raw sine series alone would not have lost.
Result
Absolutely convergent for $\alpha>1$; conditionally convergent for $\frac12<\alpha\le1$; diverges for $0<\alpha\le\frac12$: