Question

Evaluate the following limit (round off to 2 decimal places): \[ \lim_{n \to \infty} \frac{\sqrt{n+1} + \sqrt{n+2} + \cdots + \sqrt{n+n}}{\sqrt{n}} \]

Hint:

When computing limits involving sums, it's often helpful to approximate terms and use the Central Limit Theorem or other approximations for large \(n\).

Answer 1

Correct Answer: 1.1

Step 1: Simplify the expression.
We can rewrite the sum as follows: \[ S_n = \sum_{k=1}^{n} \sqrt{n+k}. \] We want to find the asymptotic behavior of this sum as \(n \to \infty\). Let's express \( \sqrt{n+k} \) as: \[ \sqrt{n+k} = \sqrt{n(1 + \frac{k}{n})} = \sqrt{n} \sqrt{1 + \frac{k}{n}}. \] Thus, the sum becomes: \[ S_n = \sum_{k=1}^{n} \sqrt{n} \sqrt{1 + \frac{k}{n}}. \] Step 2: Approximate the sum.
For large \(n\), we can use the approximation \( \sqrt{1 + \frac{k}{n}} \approx 1 + \frac{k}{2n} \) for each term in the sum. Therefore: \[ S_n \approx \sqrt{n} \sum_{k=1}^{n} \left(1 + \frac{k}{2n}\right). \] The first part of the sum is just \(n\), and the second part is the sum of the first \(n\) integers divided by \(2n\): \[ S_n \approx \sqrt{n} \left( n + \frac{1}{2n} \sum_{k=1}^{n} k \right). \] We know that \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \), so the sum becomes: \[ S_n \approx \sqrt{n} \left( n + \frac{n(n+1)}{4n} \right) = \sqrt{n} \left( n + \frac{n+1}{4} \right). \] Step 3: Take the limit.
Now, we divide the entire expression by \( \sqrt{n} \) to compute the limit: \[ \frac{S_n}{\sqrt{n}} \approx \frac{n + \frac{n+1}{4}}{\sqrt{n}} = \sqrt{n} + \frac{n+1}{4\sqrt{n}}. \] As \( n \to \infty \), the second term \( \frac{n+1}{4\sqrt{n}} \) tends to 0, so the limit is dominated by the first term: \[ \lim_{n \to \infty} \frac{S_n}{\sqrt{n}} = 2. \] Final Answer: \[ \boxed{2}. \]