Based on the general term a_n, conclude that
Based on the general term $a_n$, conclude that the following series diverges: $$ \sum_{n=0}^{\infty} \cos(n\alpha), \qquad -\frac{\pi}{2} < \alpha < \frac{\pi}{2}. $$
Explanation of Method
This mirrors Problem 14's sine series, but with cosine and a range centered at zero. The case $\alpha=0$ is immediate — every term equals $1$. For $\alpha\neq0$ in the given range, $\cos(n\alpha)$ oscillates the same way $\sin(n\alpha)$ does: periodically if $\alpha/\pi$ is rational, equidistributed if irrational — in both cases the terms never settle at zero, so the Divergence Test applies.
- Handle the special case $\alpha=0$ separately.
- Show, for $\alpha\neq0$, that $\cos(n\alpha)$ does not tend to zero via periodicity or equidistribution.
- Apply the Divergence Test to conclude in every case.
Step 1 · Identify the general term
$$ a_n = \cos(n\alpha), \qquad -\frac{\pi}{2}<\alpha<\frac{\pi}{2}. $$
Convergence would require $\lim_{n\to\infty} a_n = 0$.
Step 2 · Consider the cases for $\alpha$
Case $\alpha=0$: $a_n=\cos0=1$ for all $n$, so $\lim_{n\to\infty}a_n=1\neq0$; the series is literally $\sum 1$, which diverges.
Case $\alpha\neq0$: if $\alpha$ is a rational multiple of $\pi$, $\cos(n\alpha)$ is periodic and repeats nonzero values; if $\alpha$ is an irrational multiple of $\pi$, $\{n\alpha \bmod 2\pi\}$ is equidistributed, so $\cos(n\alpha)$ ranges throughout $[-1,1]$ infinitely often. In both cases:
- There are infinitely many $n$ with $n\alpha$ close to $0\pmod{2\pi}$, giving $\cos(n\alpha)\approx1$.
- There are infinitely many $n$ with $n\alpha$ close to $\pi\pmod{2\pi}$, giving $\cos(n\alpha)\approx-1$.
So the sequence has no single limit, and in particular does not converge to $0$.
Step 3 · Apply the divergence test
Since $\lim_{n\to\infty}\cos(n\alpha)$ is either $1$ (when $\alpha=0$) or fails to exist (when $\alpha\neq0$), it never equals $0$ on $\left(-\frac\pi2,\frac\pi2\right)$. By the Divergence Test, $\sum_{n=0}^{\infty}\cos(n\alpha)$ diverges throughout this range.
Summary Table
| Case for $\alpha$ | Behavior of $\cos(n\alpha)$ |
|---|---|
| $\alpha=0$ | Constantly $1$ |
| $\alpha\neq0$, rational multiple of $\pi$ | Periodic, repeats nonzero values |
| $\alpha\neq0$, irrational multiple of $\pi$ | Equidistributed on $[-1,1]$ |
| Conclusion | Limit never equals $0$; series diverges |
Additional Notes
- The restriction to $\left(-\frac\pi2,\frac\pi2\right)$ is not essential to the divergence conclusion itself — the same argument (aside from the $\alpha=0$ case) works for any $\alpha$ that is not a multiple of $2\pi$; the interval is likely chosen just to keep the problem's scope concrete.
- Just as with Problem 14's sine series, the partial sums $\sum_{n=0}^{N}\cos(n\alpha)$ remain bounded via the closed form $\frac{\sin\left((N+\frac12)\alpha\right)}{2\sin(\alpha/2)}+\frac12$ for $\alpha\neq0$, so the divergence here is "oscillatory," not divergence to $\pm\infty$.
- Together, Problems 14 and 15 illustrate that trigonometric series $\sum \sin(n\alpha)$ or $\sum\cos(n\alpha)$ essentially never converge in the ordinary sense (except in degenerate cases like $\alpha=0$ for sine), since their terms never decay to zero.
Result
The series diverges for every $\alpha \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ because: