Question

Given: $$ \lim_{n \to \infty} \left[ \frac{n}{(n+1)\sqrt{2n+1}} + \frac{n}{(n+2)\sqrt{2(n+2)}} + \frac{n}{(n+3)\sqrt{3(2n+3)}} + \cdots \text{(n terms)} \right] = \int_0^1 f(x)\, dx $$ Then find $ f(x) $.

• A. \( \dfrac{1}{(1 + x)\sqrt{x^2 + 2x}} \)
• B. \( \dfrac{1}{(1 + x)\sqrt{x + 2}} \)
• C. \( \dfrac{1}{(1 + x)\sqrt{x^2 + x + 1}} \)
• D. \( \dfrac{1}{(1 + x)\sqrt{x^2 - 2x}} \)

Hint:

When converting a limit-sum to an integral, use \( x = \frac{r}{n} \) and express the sum in terms of \( f(x) \cdot \frac{1}{n} \).

Answer 1

The Correct Option is A

This is a limit form of a Riemann sum: \[ \int_0^1 f(x)\, dx = \lim_{n \to \infty} \sum_{r=1}^{n} f\left(\frac{r}{n}\right)\cdot \frac{1}{n} \] Let \( x = \frac{r}{n} \), then:
- \( r = nx \)
- \( r + n = n(1 + x) \)
- \( 2r + r = 2nx + r \approx 2n x + n x = n(2x + x) = n(2x + x) \)
From the pattern, the term becomes: \[ \frac{n}{(n + r)\sqrt{2n + r}} \Rightarrow \frac{1}{(1 + x)\sqrt{x^2 + 2x}} \] So, the function under the integral is: \[ f(x) = \frac{1}{(1 + x)\sqrt{x^2 + 2x}} \]