Based on the general term a_n, conclude that
Based on the general term $a_n$, conclude that the following series diverges: $$ \sum_{n=1}^{\infty} \sin(n\alpha), \qquad 0 < \alpha < \pi. $$
Explanation of Method
Unlike a rational general term, $\sin(n\alpha)$ never settles down as $n\to\infty$ — it oscillates. To apply the Divergence Test here, it must be shown that $\sin(n\alpha)$ does not tend to $0$: either it repeats a nonzero pattern (when $\alpha/\pi$ is rational) or it is equidistributed over $[-1,1]$ (when $\alpha/\pi$ is irrational), and in both cases infinitely many terms stay bounded away from zero.
- Identify the general term and the convergence requirement $a_n \to 0$.
- Show the limit fails to exist or fails to equal zero, using periodicity or equidistribution.
- Apply the Divergence Test to conclude.
Step 1 · Identify the general term
$$ a_n = \sin(n\alpha), \qquad 0<\alpha<\pi. $$
Convergence of the series would require $\lim_{n\to\infty} a_n = 0$.
Step 2 · Show the limit does not equal zero
If $\alpha$ is a rational multiple of $\pi$: $\sin(n\alpha)$ is periodic and, for most such $\alpha$, repeats nonzero values forever. For example, with $\alpha=\frac\pi2$:
$$ \sin\!\left(\frac{n\pi}{2}\right) = 0,1,0,-1,0,1,0,-1,\ldots \quad\text{(period 4)}, $$
which oscillates and never settles at $0$.
If $\alpha$ is an irrational multiple of $\pi$: the sequence $\{n\alpha \bmod 2\pi\}$ is equidistributed on $[0,2\pi]$ (Weyl's theorem), so $\sin(n\alpha)$ takes values throughout $[-1,1]$ infinitely often — in particular, infinitely many terms stay near $1$ and infinitely many stay near $-1$, so no single limit (let alone $0$) exists.
In either case, since $0<\alpha<\pi$ gives $\sin\alpha>0$, one can find infinitely many $n$ with $n\alpha$ close to $\frac\pi2\pmod{2\pi}$ (giving $\sin(n\alpha)\approx1$) and infinitely many with $n\alpha$ close to $\frac{3\pi}2\pmod{2\pi}$ (giving $\sin(n\alpha)\approx-1$).
Step 3 · Apply the divergence test
Since $\lim_{n\to\infty}\sin(n\alpha)$ either does not exist or is bounded away from $0$, the series $\sum_{n=1}^{\infty}\sin(n\alpha)$ diverges for every $\alpha \in (0,\pi)$.
Summary Table
| Case for $\alpha$ | Behavior of $\sin(n\alpha)$ |
|---|---|
| Rational multiple of $\pi$ | Periodic, repeats nonzero values |
| Irrational multiple of $\pi$ | Equidistributed on $[-1,1]$ |
| Conclusion | Limit never equals $0$; series diverges |
Additional Notes
- The endpoints $\alpha=0$ and $\alpha=\pi$ are excluded deliberately: at $\alpha=0$, $a_n=\sin0=0$ for all $n$, so the series is trivially $\sum 0 = 0$ (convergent), and at $\alpha=\pi$, $a_n=\sin(n\pi)=0$ for all integers $n$ as well — both are degenerate cases outside the stated range.
- This proof relies on subsequence arguments rather than an explicit closed-form limit, since $\sin(n\alpha)$ genuinely has no limit as $n\to\infty$ for $\alpha\in(0,\pi)$.
- A companion result: even though $\sum\sin(n\alpha)$ diverges in the ordinary sense, its partial sums remain bounded (via the identity $\sum_{n=1}^N\sin(n\alpha) = \frac{\sin(N\alpha/2)\sin((N+1)\alpha/2)}{\sin(\alpha/2)}$), so it can still be handled by more advanced summability methods like Abel or Cesàro summation.
Result
The series diverges for every $\alpha\in(0,\pi)$ because: