To determine the values of $x$ for which the series

$\sum_{n=1}^{\infty} \frac{(3x)^n}{n^6}$

converges, we can use the ratio test or root test and analyze the resulting radius of convergence.

1. Understanding the Series Structure: The series resembles a power series in $3x$, where each term is divided by $n^6$, which decreases as $n$ increases, ensuring some form of decay. However, to determine convergence based on $x$, we focus on the term $(3x)^n$.

2. Applying the Root Test: For the series $\sum a_n$, the root test states that if

   $\lim_{n \to \infty} \sqrt[n]{|a_n|} < 1,$

   then the series converges absolutely.

   Here, each term is $a_n = \frac{(3x)^n}{n^6}$. Applying the root test:

   $\lim_{n \to \infty} \sqrt[n]{\left|\frac{(3x)^n}{n^6}\right|} = \lim_{n \to \infty} \frac{|3x|}{n^{6/n}}.$

   As $n \to \infty$, $n^{6/n} \to 1$, so this limit simplifies to $|3x|$.

   For convergence, we need:

   $|3x| < 1.$

3. Solving for $x$: $|x| < \frac{1}{3}.$

   Therefore, the interval of convergence in terms of $x$ is:

   $-\frac{1}{3} < x < \frac{1}{3}.$

4. Testing the Endpoints: We must check $x = -\frac{1}{3}$ and $x = \frac{1}{3}$ separately.

   • At $x = -\frac{1}{3}$:

     The series becomes $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^6}$, which is an alternating series with terms of the form $\frac{1}{n^6}$. Since $\frac{1}{n^6}$ converges (by the p-series test with $p = 6 > 1$), the series converges at $x = -\frac{1}{3}$.

   • At $x = \frac{1}{3}$:

     Similarly, the series becomes $\sum_{n=1}^{\infty} \frac{1}{n^6}$, which also converges by the p-series test.

5. Conclusion: The series converges for $x$ in the interval:

   $x = -\frac{1}{3}, \text{ left end included (Y)}$ to $x = \frac{1}{3}, \text{ right end included (Y)}.$

So, the answer is:

• From $x = -\frac{1}{3}$, left end