Determine whether the series is convergent or divergent.

∞

*n* = 1

1

*n*√3

convergentdivergent

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.

Learn how to determine the convergence or divergence of the series ∑(1/n√3). This problem applies the p-series test to analyze series convergence. Discover why this series diverges based on mathematical principles. Suitable for undergraduate students studying series convergence.

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