Question

Evaluate \(\lim_{n \to \infty} \frac{1}{\sqrt{n}} \left[ 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \right]\)

• A. \(\sqrt{2}\)
• B. 2
• C. \(\sqrt{2} - 1\)
• D. \(2\sqrt{2}\)

Hint:

Convert limits of sums to Riemann integrals: \(\lim_{n\to\infty} \frac{1}{n} \sum f(k/n) = \int_0^1 f(x) dx\).

Answer 1

The Correct Option is B

Step 1: Recognize as Riemann sum.
\(\lim_{n\to \infty} \frac{1}{\sqrt{n}} \sum_{k=1}^{n} \frac{1}{\sqrt{k}} = \sum_{k=1}^{n} \frac{1}{\sqrt{n}} \cdot \frac{1}{\sqrt{k}} \)

Step 2: Express in terms of \(k/n\).
\(\frac{1}{\sqrt{n}} \cdot \frac{1}{\sqrt{k}} = \frac{1}{\sqrt{n}} \cdot \frac{1}{\sqrt{(k/n)n}} = \frac{1}{\sqrt{n}} \cdot \frac{1}{\sqrt{n}} \cdot \frac{1}{\sqrt{k/n}} = \frac{1}{n} \cdot \frac{1}{\sqrt{k/n}}\)

Step 3: Define \(x = k/n\).
Then sum becomes \(\sum_{k=1}^{n} \frac{1}{n} \cdot \frac{1}{\sqrt{x}} \to \int_0^1 \frac{dx}{\sqrt{x}}\) as \(n\to\infty\)

Step 4: Integrate.
\(\int_0^1 x^{-1/2} dx = [2 \sqrt{x}]_0^1 = 2\)

Step 5: Check convergence.
Sum approximates integral, limit exists and finite

Step 6: Final conclusion.
Hence, \[ \boxed{2} \]