Using Dirichlet's test, prove the convergence of: sum
Using Dirichlet's test, prove the convergence of: $$ \sum_{n=2}^{\infty} \frac{\sin n \alpha}{\ln n}, $$ where $\alpha$ is a real parameter.
Explanation of Method
When $\alpha=k\pi$, every term vanishes and the series trivially converges. Otherwise, the classical closed-form for $\sum\sin n\alpha$ (derived from summing a geometric series of complex exponentials) shows the partial sums are bounded by $1/|\sin(\alpha/2)|$, uniformly in $N$ — combined with $1/\ln n$ decreasing to $0$, Dirichlet's test applies directly.
- Handle the trivial case $\alpha=k\pi$.
- Bound the partial sums $A_N=\sum\sin n\alpha$ using the standard closed-form identity.
- Apply Dirichlet's test with $b_n=1/\ln n$.
Step 1 · Trivial case $\alpha=k\pi$
If $\alpha=k\pi$ for integer $k$, $\sin n\alpha=\sin(nk\pi)=0$ for every $n$, so the series is identically $0$ and converges trivially.
Step 2 · Bound the partial sums for $\alpha\ne k\pi$
Using $\displaystyle\sum_{n=1}^{N}\sin n\theta = \frac{\sin\!\left(\frac{(N+1)\theta}2\right)\sin\!\left(\frac{N\theta}2\right)}{\sin(\theta/2)}$, and $\sin(\alpha/2)\ne0$ since $\alpha\ne k\pi$:
$$ |A_N| = \left|\sum_{n=2}^{N}\sin n\alpha\right| \le \frac1{|\sin(\alpha/2)|} + |\sin\alpha|, $$
a bound independent of $N$.
Step 3 · Apply Dirichlet's test
The function $f(x)=1/\ln x$ has $f'(x)=-\dfrac1{x(\ln x)^2}<0$ for $x\ge2$, so $b_n=1/\ln n$ is decreasing, and $b_n\to0$. With bounded partial sums $A_N$, all conditions of Dirichlet's test hold, so the series converges for every $\alpha\ne k\pi$ — combined with Step 1, it converges for all real $\alpha$.
Summary Table
| Condition | Conclusion |
|---|---|
| $\alpha=k\pi$ | Converges trivially (all terms $0$) |
| $\alpha\ne k\pi$ | Converges by Dirichlet's test |
Additional Notes
- The bound $|A_N|\le\frac1{|\sin(\alpha/2)|}+|\sin\alpha|$ blows up as $\alpha\to k\pi$ from either side, which is exactly consistent with the trivial case: as $\alpha$ approaches a multiple of $\pi$, the terms individually shrink to $0$, but the bound on partial sums of the un-normalized $\sin n\alpha$ becomes less tight (though the series still converges in the limit trivially).
- This closed-form bound on $\sum\sin n\theta$ is the trigonometric-series analogue of the geometric-series partial sum formula, and it recurs throughout Dirichlet-test applications with oscillatory $a_n$ (Problems 115–117 all rely on the same or a closely related identity for $\sum\cos n\theta$).
- Compared to Problem 108's $\sum(-1)^n/\ln n$ (a special case with $\alpha=\pi$ trivially reducing this to zero, or more precisely a discretely-alternating analogue), this problem generalizes the "boundedness of oscillatory partial sums" mechanism from a simple $\pm1$ pattern to a continuous family parametrized by $\alpha$.
Result
The series converges for every real $\alpha$: