To find the radius of convergence $R$ of the power series

$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n^3}$

we can apply the Root Test or Ratio Test. Let's use the Root Test for this case, which is given by:

$\frac{1}{R} = \lim_{n \to \infty} \sqrt[n]{|a_n|}$

where $a_n = \frac{(-1)^{n-1}}{n^3}$. Since $(-1)^{n-1}$ only affects the sign and not the magnitude, we have:

$|a_n| = \frac{1}{n^3}$

Now, applying the root test:

[ \frac{1}{R} = \lim_{n \to \infty} \sqrt[n]{\left|\frac{x^n}{n^3}\right|} = \lim_{n \to \infty} \sqrt[n]{\frac{|x|^n}{n^3}} ]

This simplifies to:

$\frac{1}{R} = \lim_{n \to \infty} \frac{|x|}{n^{3/n}}$

As $n \to \infty$, $n^{3/n} \to 1$, so:

$\frac{1}{R} = |x|$

Thus, the radius of convergence $R$ is:

$R = 1$

Conclusion:

The radius of convergence of the series is $R = 1$.

Would you like more details on the convergence tests or behavior at the endpoints?

Here are some related questions for deeper understanding:

1. What is the behavior of the series at $x = 1$ and $x = -1$?
2. How does the Ratio Test compare to the Root Test for this problem?
3. Can alternating series affect convergence at the endpoints?
4. How does the general term $n^3$ in the denominator influence the convergence rate?
5. What would happen if the exponent on $n$ were different, say $n^2$ or $n^4$?

Tip: The Root Test is often easier for series with powers like $n^k$, as it simplifies faster than the Ratio Test.