To determine whether the series

$\sum_{n=1}^\infty \frac{n}{n^{3/2}}$

converges or diverges using the Comparison Test, follow these steps:

Step 1: Simplify the General Term

The term of the series is: $a_n = \frac{n}{n^{3/2}} = \frac{1}{\sqrt{n}}.$

Step 2: Choose a Comparison Series

We compare this series with: $b_n = \frac{1}{n^{p}},$ where $p = 1$. The series $\sum_{n=1}^\infty b_n = \sum_{n=1}^\infty \frac{1}{n}$ is the harmonic series, which diverges.

Step 3: Apply the Limit Comparison Test

Using the Limit Comparison Test, calculate: $\lim_{n \to \infty} \frac{a_n}{b_n}.$

Substitute $a_n = \frac{1}{\sqrt{n}}$ and $b_n = \frac{1}{n}$: $\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n}}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{\sqrt{n}} = \lim_{n \to \infty} \sqrt{n}.$

As $n \to \infty$, $\sqrt{n} \to \infty$. Therefore, the limit diverges.

Step 4: Conclusion

Since the series $b_n = \frac{1}{n}$ diverges, and the Limit Comparison Test fails (the ratio does not converge to a finite, positive number), the series: $\sum_{n=1}^\infty \frac{n}{n^{3/2}}$ diverges.

Would you like further details or explanation on this? Here are some follow-up questions:

1. Why does the harmonic series diverge, and how does it relate to this test?
2. What happens if we attempt the Direct Comparison Test instead of the Limit Comparison Test?
3. Can this series be analyzed using the Integral Test? What does it show?
4. How does the exponent $3/2$ in $n^{3/2}$ influence convergence?
5. What is the role of the Limit Comparison Test in determining divergence for this series?

Tip: Always simplify the series term first to identify a suitable comparison.