Determine whether the given series are convergent.
${\sum }_{n=1}^{\infty }\frac{n^e}{n^2}$

• A. Diverges because
• B. Converges because
• C. Diverges because $p>1$
• D. Converges because $p>1$

Answer: (A)

Verified step by step guidance

1. Step 1: Recognize that the given series is \( \sum_{n=1}^{\infty} \frac{n^e}{n^2} \). This is a series where the general term is \( \frac{n^e}{n^2} \). To determine convergence, we need to analyze the behavior of the terms as \( n \to \infty \).
2. Step 2: Simplify the general term \( \frac{n^e}{n^2} \). Using the laws of exponents, rewrite it as \( n^{e-2} \). This allows us to focus on the exponent \( e-2 \) to determine the growth rate of the terms.
3. Step 3: Recall the p-series test, which states that a series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \). Here, the exponent \( e-2 \) plays the role of \( p \).
4. Step 4: Compare \( e-2 \) to 1. If \( e-2 > 1 \), the series converges. If \( e-2 \leq 1 \), the series diverges. Since \( e \) is a constant (approximately 2.718), calculate \( e-2 \) to determine whether it is greater than or less than 1.
5. Step 5: Conclude based on the value of \( e-2 \). If \( e-2 \) is less than or equal to 1, the series diverges because the terms do not decrease rapidly enough. If \( e-2 \) is greater than 1, the series converges because the terms decrease sufficiently fast.