Examine the convergence of the series: sum from n=1
Examine the convergence of the series: $$ \sum_{n=1}^{\infty} \frac{n \cos^2 \frac{n\pi}{3}}{a^n}, \qquad a>1. $$
Explanation of Method
The factor $\cos^2\frac{n\pi}3$ is periodic with period $6$, cycling through the values $\frac14,\frac14,1,\frac14,\frac14,1$, so it is always bounded between $0$ and $1$. This bound reduces the problem directly to a comparison with $\sum n/a^n$ from Problem 70, which is already known to converge for $a>1$; the root test confirms the same conclusion.
- Determine the period and bound of $\cos^2\frac{n\pi}3$.
- Compare with the known convergent series $\sum n/a^n$.
- Confirm via Cauchy's root test.
Step 1 · Period and bound of $\cos^2\frac{n\pi}3$
Since $\cos\frac{n\pi}3$ has period $6$, so does $\cos^2\frac{n\pi}3$. Direct computation for $n=1,\ldots,6$ gives the repeating pattern $\frac14,\frac14,1,\frac14,\frac14,1,\ldots$, so for every $n$: $\;0\le\cos^2\frac{n\pi}3\le1$.
Step 2 · Compare with $\sum n/a^n$
Let $u_n=\dfrac{n\cos^2\frac{n\pi}3}{a^n}$. Then:
$$ 0 \le u_n \le \frac{n}{a^n}. $$
By Problem 70, $\sum_{n=1}^\infty\dfrac n{a^n}$ converges for $a>1$. By the comparison test, $\sum u_n$ converges too.
Step 3 · Confirm via the root test
$$ \sqrt[n]{u_n} = \frac{\sqrt[n]n\cdot\sqrt[n]{\cos^2\frac{n\pi}3}}{a} \longrightarrow \frac{1\cdot1}{a} = \frac1a < 1, $$
using $\sqrt[n]n\to1$ and $0\le\sqrt[n]{\cos^2\frac{n\pi}3}\le1\to1$ whenever the factor is nonzero. By Cauchy's root test, the series converges for every $a>1$.
Summary Table
| Quantity | Value |
|---|---|
| Period of $\cos^2\frac{n\pi}3$ | $6$ |
| Bound on $\cos^2\frac{n\pi}3$ | $[0,1]$ |
| $\varlimsup\sqrt[n]{u_n}$ | $1/a<1$ |
| Conclusion | Converges for all $a>1$ |
Additional Notes
- This is a direct extension of Problem 70's $\sum n/a^n$: bounding any periodic (or merely bounded) oscillating factor between $0$ and $1$ and comparing to that base result is a technique reusable for any oscillating numerator, as also seen in Problem 81's $(-1)^n$ factor.
- Since $\cos^2$ is never negative, no absolute values are needed here, unlike series with a genuinely sign-changing factor like $(-1)^n$ or $\cos(n\alpha)$ itself (rather than its square).
- The comparison test (Step 2) and the root test (Step 3) give logically independent confirmations of convergence, illustrating that for series with a bounded oscillating factor times a rapidly-decaying base, essentially any standard test will succeed.
Result
The series converges for every $a>1$: