To determine whether the series

$\sum_{n=1}^{\infty} \left( \frac{3}{2} \right)^n$

converges or diverges, we can analyze it as a geometric series.

Step 1: Recognize the Series as a Geometric Series

The given series can be written in the form:

$\sum_{n=1}^{\infty} ar^n$

where:

• $a = \left( \frac{3}{2} \right)^1 = \frac{3}{2}$ is the first term,
• $r = \frac{3}{2}$ is the common ratio.

Step 2: Determine Convergence or Divergence

For a geometric series, the series converges if and only if the absolute value of the common ratio $|r| < 1$. In this case, we have:

$|r| = \left| \frac{3}{2} \right| = \frac{3}{2} > 1$

Since $|r| > 1$, the series diverges.

Conclusion

The series

$\sum_{n=1}^{\infty} \left( \frac{3}{2} \right)^n$

diverges because the common ratio $r = \frac{3}{2}$ is greater than 1.

Would you like more details on geometric series, or have any questions?

Here are five related questions to deepen your understanding:

1. How do you determine the convergence or divergence of a geometric series?
2. What happens if the common ratio $r$ is exactly 1 in a geometric series?
3. How can you find the sum of a convergent geometric series?
4. What is an example of a convergent geometric series, and how do you compute its sum?
5. Why does a geometric series diverge when $|r| > 1$?

Tip: For any geometric series, always check the common ratio first. If $|r| < 1$, it converges; otherwise, it diverges.