The value of `lim_(n to oo)sum_(r=1)^(n)(1)/(n) sqrt(((n+r)/(n-r)))` is :

Options

A. `(pi)/(2)-2`
B. `(pi)/(2)+2`
C. `(pi)/(2)-1`
D. `(pi)/(2)+1`

Answer

To solve the limit

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \sqrt{\frac{n+r}{n-r}}, \]

we can interpret the sum as a Riemann sum that approaches an integral as \( n \) approaches infinity. Here’s how we can derive the solution step by step:

### Step 1: Rewrite the expression inside the limit
We start with the expression:

\[ \sqrt{\frac{n+r}{n-r}}. \]

We can simplify this by factoring \( n \) out of the numerator and denominator:

\[ \sqrt{\frac{n+r}{n-r}} = \sqrt{\frac{n(1 + \frac{r}{n})}{n(1 - \frac{r}{n})}} = \sqrt{\frac{1 + \frac{r}{n}}{1 - \frac{r}{n}}}. \]

### Step 2: Substitute \( x = \frac{r}{n} \)
Let \( x = \frac{r}{n} \). Then, as \( r \) runs from \( 1 \) to \( n \), \( x \) runs from \( \frac{1}{n} \) to \( 1 \). The term \( \frac{1}{n} \) can be interpreted as \( dx \) when we take the limit as \( n \to \infty \).

### Step 3: Change the sum to an integral
The sum can now be rewritten as:

\[ \sum_{r=1}^{n} \frac{1}{n} \sqrt{\frac{1 + \frac{r}{n}}{1 - \frac{r}{n}}} \approx \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} \, dx. \]

### Step 4: Evaluate the integral
Now we need to evaluate the integral:

\[ \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} \, dx. \]

To simplify this integral, we can multiply and divide by \( \sqrt{1+x} \):

\[ \int_{0}^{1} \frac{1+x}{\sqrt{(1-x)(1+x)}} \, dx. \]

This can be split into two parts:

\[ \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \, dx + \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} \, dx. \]

### Step 5: Solve the first integral
The first integral can be recognized as:

\[ \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \, dx = \frac{\pi}{2}. \]

### Step 6: Solve the second integral
For the second integral, we can use the substitution \( u = 1 - x^2 \), which gives us:

\[ \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int_{1}^{0} u^{-1/2} \, du = \frac{1}{2}. \]

### Step 7: Combine the results
Combining both integrals, we have:

\[ \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} \, dx = \frac{\pi}{2} + \frac{1}{2}. \]

### Final Result
Thus, the final result for the limit is:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \sqrt{\frac{n+r}{n-r}} = \frac{\pi}{2} - 1. \]