Sequences And Series

Step 1: Understanding the Concept:
To determine if a sequence is convergent or divergent, we need to find the limit of the general term as approaches infinity. If the limit is a finite number, the sequence is convergent. If the limit is infinite or does not exist, the sequence is divergent.

Step 2: Key Formula or Approach:
We need to evaluate the limit:
Step 3: Detailed Explanation:
The given sequence is .
We calculate the limit as : As becomes very large, also becomes very large. Consequently, the fraction approaches 0.
Step 4: Final Answer:
Since the limit of the sequence is 0, which is a finite value, the sequence is convergent.

Step 1: Understanding the Concept:
To determine the convergence of the given series, we can use the Limit Comparison Test. This test compares the given series with a known convergent or divergent series (like a p-series) to determine its behavior.

Step 2: Key Formula or Approach:
1. Identify the general term of the series, .
2. Determine the asymptotic behavior of for large . This helps in choosing a suitable series for comparison. A p-series, , is often used, which converges for and diverges for .
3. Apply the Limit Comparison Test: If , where is a finite positive constant, then and either both converge or both diverge.

Step 3: Detailed Explanation:
Let the general term of the series be .
For large , . So, the numerator behaves like . This suggests we should compare our series with the p-series .
Let's apply the Limit Comparison Test: Divide the numerator and denominator by : Since the limit is a finite and positive number, the series converges if and only if the series converges.
The p-series converges when its exponent is greater than 1. In our case, . So, for convergence, we must have:
Step 4: Final Answer:
The series converges for , which corresponds to option (D).

Step 1: Understanding the Concept:
The expression for the sequence involves products and powers of , which is a common pattern for limits that can be evaluated by converting them into a definite integral (a Riemann sum). This is typically done by taking the logarithm of the expression. Note: The limit of as written approaches 0. However, none of the options is 0, suggesting a common exam question typo where the limit of the n-th root, , is intended. We will solve this intended problem.

Step 2: Key Formula or Approach:
1. Let .
2. Take the natural logarithm: .
3. Manipulate the expression for into the form of a Riemann sum: .
4. The limit of the Riemann sum is the definite integral: .

Step 3: Detailed Explanation:
Let's find . Taking the n-th root: Let's take the natural logarithm of this expression: We can write as . Now we take the limit as . This expression is a Riemann sum for the function over the interval . We evaluate the integral using integration by parts, . Let and . Then and . Evaluating the definite integral: So, . Therefore, .
Step 4: Final Answer:
Assuming the question intended to ask for the limit of , the value is .