To determine the convergence or divergence of the series

[ \sum_{n=1}^{\infty} \frac{5^{n+1}}{3^n - 1}, ]

we can use the Limit Comparison Test. This requires us to compare the given series with a known reference series.

Step 1: Simplification

First, let's simplify the term in the series:

[ \frac{5^{n+1}}{3^n - 1} = \frac{5 \cdot 5^n}{3^n - 1}. ]

As (n) becomes very large, the (-1) in the denominator becomes negligible compared to (3^n). Therefore, we can approximate:

[ 3^n - 1 \sim 3^n \quad \text{(as } n \to \infty \text{)}. ]

Consequently,

[ \frac{5^{n+1}}{3^n - 1} \sim \frac{5 \cdot 5^n}{3^n} = 5 \frac{5^n}{3^n} = 5 \left( \frac{5}{3} \right)^n. ]

Step 2: Choosing a Comparison Series

Now we can choose the series

[ \sum_{n=1}^{\infty} \left( \frac{5}{3} \right)^n ]

as our reference series. This is a geometric series with ratio

[ r = \frac{5}{3} > 1. ]

Step 3: Evaluate Convergence of the Comparison Series

Since (r > 1), the series

[ \sum_{n=1}^{\infty} \left( \frac{5}{3} \right)^n ]

diverges.

Step 4: Applying the Limit Comparison Test

Now we apply the Limit Comparison Test. We will find the limit:

[ L = \lim_{n \to \infty} \frac{\frac{5^{n+1}}{3^n - 1}}{5 \left( \frac{5}{3} \right)^n}. ]

Calculating this limit:

[ L = \lim_{n \to \infty} \frac{5^{n+1}}{3^n - 1} \cdot \frac{3^n}{5^n} = \lim_{n \to \infty} \frac{5 \cdot 5^n}{3^n - 1} \cdot \frac{3^n}{5^n} = \lim_{n \to \infty} \frac{5 \cdot 3^n}{3^n - 1}. ]

Since (3^n - 1 \sim 3^n) for large (n):

[ L = \lim_{n \to \infty} \frac{5 \cdot 3^n}{3^n} = 5. ]

Conclusion

Since (L) is a positive finite number (specifically, (5)), the Limit Comparison Test tells us that both series either converge or diverge together.

Since the comparison series

[ \sum_{n=1}^{\infty} \left( \frac{5}{3} \right)^n ]

diverges, it follows that the original series

[ \sum_{n=1}^{\infty} \frac{5^{n+1}}{3^n - 1} ] also diverges.

Thus, the series

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

diverges.