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The expression provided is:

\[
\sum_{n=2}^{\infty} \frac{5^{n+1}}{2 \cdot 3^n}
\]

To solve this infinite series, let's follow these steps:

1. **Factor out the constant term**: The factor \( 5^{n+1} \) can be rewritten as \( 5 \cdot 5^n \), and the denominator \( 2 \cdot 3^n \) remains. So the expression becomes:

\[
5 \sum_{n=2}^{\infty} \frac{5^n}{2 \cdot 3^n}
\]

2. **Rewrite the fraction**: The expression inside the summation becomes:

\[
\frac{5}{2} \sum_{n=2}^{\infty} \left( \frac{5}{3} \right)^n
\]

3. **Simplify the summation**: This is now a geometric series of the form:

\[
\sum_{n=2}^{\infty} r^n
\]

Where \( r = \frac{5}{3} \). For a geometric series starting at \( n = 2 \), the sum is:

\[
\sum_{n=2}^{\infty} r^n = \frac{r^2}{1 - r}
\]

However, this series only converges if \( |r| < 1 \), but since \( r = \frac{5}{3} \) is greater than 1, the series **diverges**.

Thus, the infinite series does not have a finite sum because the common ratio is greater than 1, leading to divergence.

Would you like further details or clarifications?

Here are 5 related questions to consider:
1. How do you determine if a geometric series converges or diverges?
2. Can you apply this method to other types of summations with different ratios?
3. What happens if the ratio \( r \) is less than 1?
4. How would the summation change if it started at \( n = 0 \)?
5. Could this series be modified to achieve convergence?

**Tip**: For a geometric series to converge, the absolute value of the ratio must be less than 1. Otherwise, it will diverge.

Evaluate the infinite series: ∑ from n=2 to ∞ (5^(n+1)) / (2 * 3^n)

Solve the infinite series ∑ (5^(n+1)) / (2 * 3^n) starting at n=2. Learn how to apply geometric series concepts, factor exponents, and determine series divergence when the ratio exceeds 1. Suitable for advanced math students.

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