Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty}(-1)^{n-1} \frac{2^n \sin^{2n} x}{n}. $$
Explanation of Method
Writing $r=2\sin^2x\in[0,2]$ collapses the series to the standard parametrized form $\sum(-1)^{n-1}r^n/n$, whose behavior is completely classified: the ratio test gives limit $r$ for the absolute series, so $r<1$ means absolute convergence, $r>1$ means divergence (terms don't vanish), and the single boundary case $r=1$ reduces exactly to the classical alternating harmonic series.
- Substitute $r=2\sin^2x$ and identify the resulting series.
- Apply the ratio test to the absolute series $\sum r^n/n$.
- Resolve each regime of $r$, including the boundary $r=1$.
Step 1 · Substitute and identify the series
Since $2^n\sin^{2n}x=(2\sin^2x)^n=r^n$ with $r=2\sin^2x\in[0,2]$ (as $0\le\sin^2x\le1$):
$$ a_n = (-1)^{n-1}\frac{r^n}n. $$
Step 2 · Ratio test on the absolute series
$$ \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} = \lim_{n\to\infty} r\cdot\frac n{n+1} = r. $$
Step 3 · Resolve each regime
If $r<1$ (i.e. $|\sin x|<\frac1{\sqrt2}$): absolute convergence. If $r>1$ (i.e. $|\sin x|>\frac1{\sqrt2}$): $r^n/n\to\infty$, terms don't vanish, series diverges. If $r=1$ (i.e. $|\sin x|=\frac1{\sqrt2}$): the series is exactly $\sum(-1)^{n-1}/n=\ln2$, convergent by Leibniz's test but $\sum1/n$ diverges — conditional convergence.
Summary Table
| Condition | Conclusion |
|---|---|
| $|\sin x|<1/\sqrt2$ | Absolutely convergent |
| $|\sin x|=1/\sqrt2$ | Conditionally convergent |
| $|\sin x|>1/\sqrt2$ | Diverges |
Additional Notes
- The critical value $|\sin x|=\frac1{\sqrt2}$ corresponds to $x=\frac\pi4,\frac{3\pi}4,\ldots$ — angles where the trigonometric substitution lands exactly on the boundary of the geometric-type series' convergence disk, a direct trigonometric analogue of a power series' radius of convergence.
- This problem is structurally the trigonometric twin of a power series $\sum(-1)^{n-1}z^n/n=\ln(1+z)$ evaluated on its boundary circle $|z|=1$: convergence at $z=1$ (giving $\ln2$) but divergence at $z=-1$ — here the "radius" condition $r=1$ translates to the specific angle where $\sin^2x=\frac12$.
- Every value of $x$ falls into exactly one of the three regimes depending only on $|\sin x|$, making this among the cleanest possible trigonometric convergence classifications, directly comparable in structure to Problem 132's $\cos(n\pi/4)/n^a$ but parametrized by $x$ rather than by a power $a$.
Result
Convergence depends entirely on $|\sin x|$: