Question

Evaluate the limit \[ \lim_{n \to \infty} \frac{1}{\sqrt{n}} \left( \frac{1}{\sqrt{3} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{9}} + \cdots + \frac{1}{\sqrt{3n} + \sqrt{3n + 3}} \right). \]

• A. \( 1 + \sqrt{3} \)
• B. \( \frac{1}{\sqrt{3}} \)
• C. \( \sqrt{3} \)
• D. \( \frac{1}{1 + \sqrt{3}} \)

Hint:

For telescoping series, carefully simplify the terms and look for cancellations to compute the sum.

Answer 1

The Correct Option is B

Step 1: Simplifying the terms.
The general term of the sum can be written as: \[ \frac{1}{\sqrt{3n} + \sqrt{3n+3}}. \] By rationalizing the denominator and simplifying, we get: \[ \frac{1}{\sqrt{3n} + \sqrt{3n+3}} = \frac{\sqrt{3n+3} - \sqrt{3n}}{(3n+3) - 3n} = \sqrt{3n+3} - \sqrt{3n}. \]
Step 2: Summing the series.
We now sum the series: \[ S_n = \sum_{k=1}^{n} \left( \sqrt{3k+3} - \sqrt{3k} \right). \] This is a telescoping series, and after canceling terms, we get: \[ S_n = \sqrt{3n+3} - \sqrt{3}. \]
Step 3: Taking the limit.
Now, dividing by \( \sqrt{n} \) and taking the limit as \( n \to \infty \), we find that the limit of the sum is \( \frac{1}{\sqrt{3}} \).
Step 4: Conclusion.
Thus, the correct answer is (B) \( \frac{1}{\sqrt{3}} \).