Evaluate `S=sum_(r=0)^(n-1)(1)/(sqrt(4n^(2)-r^(2)))as n rarrinfty`.

Answer

To solve the problem \( S = \sum_{r=0}^{n-1} \frac{1}{\sqrt{4n^2 - r^2}} \) as \( n \to \infty \), we will follow these steps:

### Step 1: Rewrite the Summation
We start with the expression:
\[ S = \sum_{r=0}^{n-1} \frac{1}{\sqrt{4n^2 - r^2}} \]
As \( n \to \infty \), we can interpret this summation as a Riemann sum. To do this, we will express \( r \) in terms of \( n \).

### Step 2: Change of Variables
Let \( x = \frac{r}{n} \). Then, \( r = nx \) and as \( r \) goes from \( 0 \) to \( n-1 \), \( x \) goes from \( 0 \) to \( 1 - \frac{1}{n} \). The increment \( dr \) can be approximated as \( n \, dx \).

Thus, we rewrite the summation:
\[ S \approx \sum_{r=0}^{n-1} \frac{1}{\sqrt{4n^2 - (nx)^2}} \cdot \frac{1}{n} \]
This becomes:
\[ S \approx \int_{0}^{1} \frac{1}{\sqrt{4n^2 - n^2x^2}} \, dx \]

### Step 3: Simplify the Integral
We can factor out \( n^2 \) from the square root:
\[ S \approx \int_{0}^{1} \frac{1}{\sqrt{n^2(4 - x^2)}} \, dx = \frac{1}{n} \int_{0}^{1} \frac{1}{\sqrt{4 - x^2}} \, dx \]

### Step 4: Evaluate the Integral
The integral \( \int \frac{1}{\sqrt{4 - x^2}} \, dx \) can be evaluated using the substitution \( x = 2 \sin(\theta) \):
\[ dx = 2 \cos(\theta) \, d\theta \]
The limits change as follows:
- When \( x = 0 \), \( \theta = 0 \)
- When \( x = 1 \), \( \theta = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \)

Thus, the integral becomes:
\[ \int_{0}^{\frac{\pi}{6}} \frac{2 \cos(\theta)}{\sqrt{4 - 4\sin^2(\theta)}} \, d\theta = \int_{0}^{\frac{\pi}{6}} \frac{2 \cos(\theta)}{2 \cos(\theta)} \, d\theta = \int_{0}^{\frac{\pi}{6}} 1 \, d\theta = \frac{\pi}{6} \]

### Step 5: Final Calculation
Putting everything together, we have:
\[ S \approx \frac{1}{n} \cdot \frac{\pi}{6} \]
As \( n \to \infty \), the term \( \frac{1}{n} \) approaches \( 0 \), and thus:
\[ S \to \frac{\pi}{6} \]

### Conclusion
Therefore, the final result is:
\[ \boxed{\frac{\pi}{6}} \]