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: Reshape the sum to look like a Riemann sum.
Write the bracket terms with a common $\sqrt{n}$. Each term $\frac{1}{\sqrt{k}}$ can be written as $\frac{1}{\sqrt{n}}\cdot\frac{1}{\sqrt{k/n}}$. So the whole expression is $\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}} = \frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{k/n}}$.

Step 2: Identify the matching function.
The piece $\frac{1}{\sqrt{k/n}}$ is $g\!\left(\frac{k}{n}\right)$ where $g(x) = \frac{1}{\sqrt{x}}$. The prefactor $\frac{1}{n}$ is the width of each strip.

Step 3: Recall the Riemann sum to integral rule.
As $n \to \infty$, $\frac{1}{n}\sum_{k=1}^{n} g\!\left(\frac{k}{n}\right) \to \int_0^1 g(x)\,dx$.

Step 4: Set up the integral.
The limit equals $\int_0^1 \frac{1}{\sqrt{x}}\,dx = \int_0^1 x^{-1/2}\,dx$.

Step 5: Integrate.
$\int x^{-1/2}\,dx = 2\sqrt{x}$. Evaluated from $0$ to $1$: $2\sqrt{1} - 2\sqrt{0} = 2 - 0 = 2$.

Step 6: State the answer.
The improper integral converges, so the limit equals $2$.
\[ \boxed{2} \]