Question

The limit \[ \lim_{n \to \infty} \frac{1}{\sqrt{n}} \left[ 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} \right] \]

• A. equals 2
• B. equals 1
• C. equals 0
• D. does not exist

Hint:

An alternative way to solve this is using the Squeeze Theorem by bounding the sum with the integral:
\[ \int_{1}^{n+1} \frac{1}{\sqrt{x}} \, dx \lt \sum_{i=1}^{n} \frac{1}{\sqrt{i}} \lt 1 + \int_{1}^{n} \frac{1}{\sqrt{x}} \, dx \]
Evaluating and dividing by $\sqrt{n}$ easily shows both sides approach 2.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks for the limit of a sum of square-root reciprocals scaled by $\frac{1}{\sqrt{n}}$ as $n$ approaches infinity.

Step 2: Key Formula or Approach:
We can evaluate this limit using the Riemann Sum approximation of definite integrals:
\[ \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} g\left(\frac{i}{n}\right) = \int_{0}^{1} g(x) \, dx \]

Step 3: Detailed Explanation:

• Let us rewrite the given expression inside the limit:
\[ L = \lim_{n \to \infty} \frac{1}{\sqrt{n}} \sum_{i=1}^{n} \frac{1}{\sqrt{i}} \]

• We can rewrite this by multiplying and dividing by $\sqrt{n}$ to bring it into the standard Riemann sum form:
\[ L = \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n} \cdot \sqrt{n}} \sum_{i=1}^{n} \frac{1}{\sqrt{i}} \]
\[ L = \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \frac{\sqrt{n}}{\sqrt{i}} \]
\[ L = \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^{n} \frac{1}{\sqrt{\frac{i}{n}}} \]

• This matches the standard Riemann sum structure where $g(x) = \frac{1}{\sqrt{x}}$:
\[ L = \int_{0}^{1} \frac{1}{\sqrt{x}} \, dx \]

• Let us evaluate this definite integral:
\[ I = \int_{0}^{1} x^{-1/2} \, dx = \left[ 2x^{1/2} \right]_{0}^{1} = 2(1) - 2(0) = 2 \]



Step 4: Final Answer:
The limit equals 2.