Determine the region of absolute and conditional
Determine the region of absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac{2^n\sin^nx}{n^2}. $$
Explanation of Method
With $y=2\sin x\in[-2,2]$, this is the classical series $\sum y^n/n^2$, resolved by the root test with threshold $|y|<1$, i.e. $|\sin x|<\frac12$. Because the boundary series $\sum1/n^2$ converges absolutely (a $p$-series with $p=2>1$), the boundary points $|\sin x|=\frac12$ are included in the region of absolute convergence, leaving no room for conditional convergence anywhere.
- Substitute $y=2\sin x$ and apply the root test.
- Translate the threshold $|y|<1$ into a condition on $\sin x$.
- Check the boundary $|\sin x|=\frac12$ directly, confirming absolute convergence there too.
Step 1 · Root test
With $y=2\sin x$: $\sqrt[n]{|y|^n/n^2}=|y|/n^{2/n}\to|y|$ (since $n^{2/n}\to1$). So $\sum y^n/n^2$ converges for $|y|<1$, diverges for $|y|>1$.
Step 2 · Translate to $\sin x$
$|y|<1\iff|2\sin x|<1\iff|\sin x|<\dfrac12$.
Step 3 · Check the boundary
At $|\sin x|=\frac12$: $|y|=1$, and $\sum1/n^2$ converges absolutely ($p=2>1$), regardless of sign — whether $\sin x=\frac12$ (giving $\sum1/n^2$) or $\sin x=-\frac12$ (giving $\sum(-1)^n/n^2$, absolutely convergent by comparison). So both boundary cases are absolutely convergent.
Summary Table
| Condition | Conclusion |
|---|---|
| $|\sin x|\le\frac12$ | Absolutely convergent |
| $|\sin x|>\frac12$ | Diverges |
Additional Notes
- This problem is the $1/n^2$-weighted counterpart to Problem 128's $1/n$-weighted trigonometric series $\sum(-1)^{n-1}(2\sin^2x)^n/n$: the stronger decay $1/n^2$ here (versus $1/n$) is exactly what pushes the boundary case into absolute rather than conditional convergence.
- The convergence region $|\sin x|\le\frac12$ corresponds to $x$ lying in specific unions of closed intervals around each multiple of $\pi$, e.g. $\left[-\frac\pi6,\frac\pi6\right]$ (mod $2\pi$) and its reflections, unlike Problem 128's open-interval-plus-single-point structure.
- Since the underlying series $\sum y^n/n^2$ is absolutely convergent on its entire closed disk $|y|\le1$ (a general fact for $p$-series exponents $>1$), any trigonometric substitution into this template automatically inherits full boundary inclusion — a useful shortcut compared to case-by-case boundary analysis.
Result
Converges absolutely for $|\sin x|\le\frac12$; diverges otherwise: