Determine the convergence or divergence of the series. 
${\sum }_{n=1}^{\infty }\frac{5^{n-1}}{2^n}$

• A. Inconclusive
• B. Diverges
• C. Converges

Answer: (B)

Verified step by step guidance

1. Step 1: Recognize that the given series is a geometric series. A geometric series has the form \( \sum_{n=1}^{\infty} ar^{n-1} \), where \( a \) is the first term and \( r \) is the common ratio.
2. Step 2: Rewrite the given series \( \sum_{n=1}^{\infty} \frac{5^{n-1}}{2^{n}} \) to match the geometric series form. Factor out \( \frac{1}{2} \) from the denominator to express the series as \( \sum_{n=1}^{\infty} \left( \frac{5}{2} \right)^{n-1} \cdot \frac{1}{2} \).
3. Step 3: Identify the first term \( a \) and the common ratio \( r \). Here, \( a = \frac{1}{2} \) and \( r = \frac{5}{2} \).
4. Step 4: Recall the convergence criterion for a geometric series: A geometric series converges if and only if \( |r| < 1 \). Check whether \( |r| = \left| \frac{5}{2} \right| \) satisfies this condition.
5. Step 5: Conclude whether the series converges or diverges based on the value of \( |r| \). If \( |r| \geq 1 \), the series diverges; otherwise, it converges.