Determine the region of absolute and conditional
Determine the region of absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} n\sqrt[3]{\sin^nx}. $$
Explanation of Method
The real cube root is well-defined for negative arguments, so $\sqrt[3]{\sin^nx}=\left(\sqrt[3]{\sin x}\right)^n=r^n$ where $r=\sqrt[3]{\sin x}$ is a real number with $|r|=|\sin x|^{1/3}\le1$. This reduces the whole problem to the classical series $\sum nr^n$, resolved by the ratio (or root) test, with the only possible boundary values $r=\pm1$ occurring exactly when $\sin x=\pm1$.
- Rewrite the term as $nr^n$ with $r=\sqrt[3]{\sin x}$, the real cube root.
- Apply the ratio test to $\sum nr^n$.
- Check the boundary cases $\sin x=\pm1$.
Step 1 · Rewrite via the real cube root
Since cube-rooting is an odd, monotonic bijection on $\mathbb R$: $\sqrt[3]{\sin^nx}=\left(\sqrt[3]{\sin x}\right)^n$. Let $r=\sqrt[3]{\sin x}$; since $-1\le\sin x\le1$, also $-1\le r\le1$. The term is $a_n=nr^n$.
Step 2 · Ratio test on $\sum nr^n$
$\left|\dfrac{a_{n+1}}{a_n}\right|=|r|\dfrac{n+1}n\to|r|$, so the series converges absolutely for $|r|<1$ (i.e. $|\sin x|<1$) and diverges for $|r|>1$ (impossible here, since $|r|\le1$ always).
Step 3 · Boundary cases $|r|=1$
$|r|=1$ occurs exactly when $\sin x=\pm1$, i.e. $x=\frac\pi2+k\pi$. At $\sin x=1$ ($r=1$): $\sum n$ diverges (terms $\to\infty$). At $\sin x=-1$ ($r=-1$): $\sum n(-1)^n$ diverges, since the terms $\pm n\not\to0$.
Summary Table
| Condition | Conclusion |
|---|---|
| $|\sin x|<1$ (i.e. $x\ne\frac\pi2+k\pi$) | Absolutely convergent |
| $\sin x=\pm1$ (i.e. $x=\frac\pi2+k\pi$) | Diverges |
Additional Notes
- Because the cube root is defined and continuous on all of $\mathbb R$ (unlike a square or general fractional root), no complex-number subtleties arise here: $r=\sqrt[3]{\sin x}$ is simply a real number in $[-1,1]$ for every $x$, and the whole problem reduces cleanly to the elementary series $\sum nr^n$.
- Since $\sum nr^n$ has coefficients that grow polynomially (like Problem 164's $\sum n/x^n$), there is no region of conditional convergence — wherever the series converges, it does so absolutely.
- The excluded points $x=\frac\pi2+k\pi$ are exactly the values where $|\sin x|$ reaches its maximum $1$, a boundary phenomenon directly analogous to Problem 128's critical angle $|\sin x|=1/\sqrt2$, but here occurring at the extreme values of sine rather than an interior threshold.
Result
Converges absolutely except at the points where $|\sin x|=1$: