Let's analyze the given series:

$\sum_{n=1}^{\infty} \frac{1}{n\sqrt{3}}$

This series is similar to a p-series, which has the general form:

$\sum_{n=1}^{\infty} \frac{1}{n^p}$

A p-series converges if $p > 1$ and diverges if $p \leq 1$.

For the series given, let's rewrite it slightly:

$\sum_{n=1}^{\infty} \frac{1}{n\sqrt{3}} = \sum_{n=1}^{\infty} \frac{1}{n^{1/\sqrt{3}}}$

Here, the exponent $p = \frac{1}{\sqrt{3}}$.

Since $\sqrt{3}$ is approximately 1.732, $\frac{1}{\sqrt{3}}$ is approximately 0.577. This value is less than 1.

Because $p \leq 1$, the series diverges by the p-series test.

Answer: The series is divergent.

Would you like more details, or do you have any questions? Here are some related questions for further exploration:

1. What is the p-series test, and how is it applied in different scenarios?
2. Can you determine the convergence of a series using the comparison test?
3. What is the harmonic series, and why does it diverge?
4. How does the ratio test help in determining the convergence of a series?
5. What are some examples of convergent p-series?

Tip: When dealing with series, always check if a simpler known test (like the p-series test) can be applied to quickly determine convergence or divergence.