Question

If \[ \lim_{n\to\infty} \left[ \frac{n}{(n+1)\sqrt{2n+1}} + \frac{n}{(n+2)\sqrt{2(2n+2)}} + \frac{n}{(n+3)\sqrt{3(2n+3)}} +\cdots+n\ \text{terms} \right] = \int_0^1 f(x)\,dx, \] then \(f(x)=\)

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

Hint:

To convert a limit of sums into a definite integral, write \[ x=\frac{r}{n} \] and express the sum in the form \[ \sum \frac1n f\left(\frac{r}{n}\right). \]

Answer 1

The Correct Option is A

Step 1: Write the general term.
The general term of the sum is \[ \frac{n}{(n+r)\sqrt{r(2n+r)}}, \qquad r=1,2,\dots,n \] Let \[ x=\frac{r}{n} \] Then, \[ r=nx \]

Step 2: Rewrite the term in terms of \(x\).
Now, \[ n+r=n(1+x) \] Also, \[ r(2n+r) = (nx)(2n+nx) \] \[ =n^2x(2+x) \] Hence, \[ \sqrt{r(2n+r)} = n\sqrt{x(2+x)} \] Therefore, \[ \frac{n}{(n+r)\sqrt{r(2n+r)}} = \frac{n}{n(1+x)\cdot n\sqrt{x(2+x)}} \] \[ = \frac{1}{n(1+x)\sqrt{x(2+x)}} \]

Step 3: Identify the Riemann sum.
Thus, \[ \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{1}{(1+x)\sqrt{x(2+x)}} \] As \[ n\to\infty, \] this becomes \[ \int_0^1 \frac{1}{(1+x)\sqrt{x(2+x)}}dx \] Since \[ x(2+x)=x^2+2x, \] we get \[ f(x)=\frac{1}{(1+x)\sqrt{x^2+2x}} \]

Step 4: Final conclusion.
Therefore, \[ \boxed{ \frac{1}{(1+x)\sqrt{x^2+2x}} } \]