Question

The value of the limit \[ \lim_{n \to \infty} \left( \left(1 + \frac{1}{n}\right) \left(1 + \frac{2}{n}\right) \cdots \left(1 + \frac{n}{n}\right) \right)^{\frac{1}{n}} \] is equal to:

• A. \( e \)
• B. \( \frac{1}{e} \)
• C. \( \frac{3}{e} \)
• D. \( \frac{4}{e} \)

Hint:

When a product involves terms like \((1 + \frac{k}{n})\), converting it to a Riemann sum via logarithms often simplifies the problem to an integral form.

Answer 1

The Correct Option is D

Step 1: Rewrite the product.
Let \[ L = \lim_{n \to \infty} \left( \prod_{k=1}^{n} \left(1 + \frac{k}{n}\right) \right)^{\frac{1}{n}} \] Take logarithm on both sides: \[ \ln L = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln\left(1 + \frac{k}{n}\right) \]
Step 2: Express as a Riemann sum.
\[ \ln L = \int_{0}^{1} \ln(1 + x) \, dx \]
Step 3: Evaluate the integral.
Using integration by parts: \[ \int \ln(1 + x) \, dx = (1 + x)\ln(1 + x) - x + C \] Substitute limits 0 to 1: \[ \int_{0}^{1} \ln(1 + x) \, dx = [2\ln2 - 1] \]
Step 4: Find the limit.
\[ \ln L = 2\ln2 - 1 \Rightarrow L = e^{2\ln2 - 1} = \frac{4}{e} \] Final Answer: \[ \boxed{\frac{4}{e}} \]