Question

Evaluate the limit: \[ \lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2}} + \frac{1}{\sqrt{n^2 - 1}} + \dots + \frac{1}{\sqrt{n^2 - (n-1)^2}} \right). \]

• A. \( 2\sqrt{\pi} \)
• B. \( \frac{2}{\sqrt{\pi}} \)
• C. \( \frac{\pi}{2} \)
• D. \( \frac{3\pi}{2} \)

Hint:

For summations involving terms of the form \( \frac{1}{\sqrt{n^2 - k^2}} \), use the integral approximation: \[ \sum \approx \int \frac{dx}{\sqrt{n^2 - x^2}}. \] Recognizing the standard integral \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left( \frac{x}{a} \right) \) helps in evaluating the limit.

Answer 1

The Correct Option is C

Step 1: Expressing the Summation as an Integral The given sum can be rewritten as: \[ S_n = \sum_{k=1}^{n} \frac{1}{\sqrt{n^2 - k^2}}. \] For large values of \( n \), we approximate the sum using an integral: \[ S_n \approx \int_0^n \frac{dx}{\sqrt{n^2 - x^2}}. \] Using the standard integral result: \[ \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left( \frac{x}{a} \right), \] we evaluate: \[ S_n \approx \int_0^n \frac{dx}{\sqrt{n^2 - x^2}}. \]
Step 2: Evaluating the Integral Using the substitution \( x = n \sin \theta \), so that \( dx = n \cos \theta d\theta \), we transform the integral into: \[ I = \int_0^n \frac{dx}{\sqrt{n^2 - x^2}}. \] Since \( x = n \sin \theta \), \[ dx = n \cos \theta d\theta. \] Thus, the integral simplifies to: \[ \int_0^{\frac{\pi}{2}} d\theta = \frac{\pi}{2}. \]
Step 3: Conclusion Taking the limit as \( n \to \infty \), we obtain: \[ \lim_{n \to \infty} S_n = \frac{\pi}{2}. \] % Final Answer Thus, the correct answer is option (3): \( \frac{\pi}{2} \).

Answer 2

To evaluate the limit: \[ \lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2}} + \frac{1}{\sqrt{n^2 - 1}} + \dots + \frac{1}{\sqrt{n^2 - (n-1)^2}} \right), \] we first rewrite the expression as a sum: \[ \sum_{k=0}^{n-1} \frac{1}{\sqrt{n^2 - k^2}}. \] Consider the substitution \( x = \frac{k}{n} \), which implies \( k = nx \) and as \( n \to \infty \), \(\frac{1}{n}\) becomes the differential \(dx\). The limit converts the discrete sum to an integral: \[ \lim_{n \to \infty} \sum_{k=0}^{n-1} \frac{1}{\sqrt{n^2 - k^2}} \cdot \frac{1}{n} = \int_{0}^{1} \frac{1}{\sqrt{1 - x^2}} \, dx. \] The integral \(\int_{0}^{1} \frac{1}{\sqrt{1 - x^2}} \, dx\) is recognized as the integral representing half the area of the unit circle, which is the arcsine function: \[ \int_{0}^{x} \frac{1}{\sqrt{1 - u^2}} \, du = \arcsin(x) \bigg|_{0}^{1} = \arcsin(1) - \arcsin(0). \] Calculating gives: \[ \arcsin(1) = \frac{\pi}{2}, \quad \arcsin(0) = 0. \] Thus, the integral evaluates to: \[ \frac{\pi}{2} - 0 = \frac{\pi}{2}. \] Therefore, the limit is: \[ \boxed{\frac{\pi}{2}}. \]