Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=2}^{\infty} \frac{\sin \frac{n\pi}{12}}{\ln n}. $$
Explanation of Method
Since $\left|\sin\frac{n\pi}{12}\right|$ is bounded away from zero for all but a sparse set of indices (multiples of $12$), the absolute series compares to (a constant multiple of) the divergent $\sum1/\ln n$ from Problem 108. But Dirichlet's test still applies directly: the closed-form sine-sum bounds the partial sums of $\sin\frac{n\pi}{12}$ uniformly, while $1/\ln n$ decreases to $0$.
- Show $\sum\left|\sin\frac{n\pi}{12}\right|/\ln n$ diverges by comparison with $\sum1/\ln n$.
- Bound the partial sums of $\sin\frac{n\pi}{12}$ via the closed-form identity.
- Apply Dirichlet's test to conclude conditional convergence.
Step 1 · Rule out absolute convergence
Since $\sin\frac{n\pi}{12}=0$ only when $12\mid n$, for the remaining $\frac{11}{12}$ of indices $\left|\sin\frac{n\pi}{12}\right|\ge c$ for some fixed $c>0$ (the smallest nonzero value in the period-$24$ pattern). So $\sum\left|\sin\frac{n\pi}{12}\right|/\ln n \ge c\sum_{n\not\equiv0(12)}1/\ln n$, and since $\sum1/\ln n$ diverges (Problem 108), so does this comparison series. No absolute convergence.
Step 2 · Bound the partial sums
Using $\sum_{n=1}^N\sin n\theta=\dfrac{\sin\left(\frac{(N+1)\theta}2\right)\sin\left(\frac{N\theta}2\right)}{\sin(\theta/2)}$ with $\theta=\pi/12$, and $\sin(\pi/24)\ne0$:
$$ |A_N| \le \frac1{\sin(\pi/24)} + \left|\sin\frac\pi{12}\right| \approx 7.66+1 < 9. $$
Step 3 · Apply Dirichlet's test
With $b_n=\frac1{\ln n}$ decreasing to $0$ for $n\ge2$, and bounded partial sums from Step 2, Dirichlet's test gives convergence. Combined with Step 1, the series converges conditionally.
Summary Table
| Test | Result |
|---|---|
| $\sum|\sin(n\pi/12)|/\ln n$ | Diverges (comparison with $\sum1/\ln n$) |
| Dirichlet's test | All conditions satisfied |
| Conclusion | Conditionally convergent |
Additional Notes
- This problem directly generalizes Problem 108's $\sum(-1)^n/\ln n$: instead of the simple period-2 sign pattern $(-1)^n$, the oscillating factor here is $\sin\frac{n\pi}{12}$ with period $24$, but the underlying convergence mechanism (Dirichlet's test against a divergent-in-absolute-value comparison) is identical.
- The choice of period $12$ (rather than, say, the period-$8$ pattern of Problems 116 or 132) doesn't change the qualitative outcome — any nonzero angle $\theta=\pi/k$ for integer $k$ produces a bounded partial-sum constant $1/\sin(\theta/2)$, just with a different numerical value.
- Since $1/\ln n$ decays even slower than $1/n^a$ for any $a>0$, this is among the "weakest" possible decreasing multipliers for which Dirichlet's test still guarantees convergence — a reminder that the test's power comes from boundedness of oscillation, not from the rate of decay of the multiplier.
Result
The series converges conditionally but not absolutely: