Evaluate
`lim _( n to oo) sum_( r =1) ^(n -1) (1)/(sqrt(n ^(2) -r ^(2)))`

Answer

To evaluate the limit

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

we can follow these steps:

### Step 1: Rewrite the Sum
First, we can factor out \(n\) from the square root in the denominator:

\[ \sqrt{n^2 - r^2} = n \sqrt{1 - \left(\frac{r}{n}\right)^2}. \]

Thus, we can rewrite the sum as:

\[ \sum_{r=1}^{n-1} \frac{1}{\sqrt{n^2 - r^2}} = \sum_{r=1}^{n-1} \frac{1}{n \sqrt{1 - \left(\frac{r}{n}\right)^2}}. \]

### Step 2: Change the Variable
Next, we change the variable by letting \(x = \frac{r}{n}\). Then, \(r = nx\) and as \(r\) goes from \(1\) to \(n-1\), \(x\) goes from \(\frac{1}{n}\) to \(\frac{n-1}{n}\). The increment \(dr\) becomes \(n \, dx\).

Thus, we can express the sum as:

\[ \sum_{r=1}^{n-1} \frac{1}{n \sqrt{1 - x^2}} \cdot n \, dx = \sum_{r=1}^{n-1} \frac{1}{\sqrt{1 - x^2}} \, dx. \]

### Step 3: Convert the Sum to an Integral
As \(n\) approaches infinity, the sum can be approximated by an integral:

\[ \lim_{n \to \infty} \sum_{r=1}^{n-1} \frac{1}{\sqrt{1 - x^2}} \cdot \frac{1}{n} \approx \int_0^1 \frac{1}{\sqrt{1 - x^2}} \, dx. \]

### Step 4: Evaluate the Integral
The integral

\[ \int_0^1 \frac{1}{\sqrt{1 - x^2}} \, dx \]

is a standard integral that evaluates to:

\[ \left[ \sin^{-1}(x) \right]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \]

### Step 5: Conclusion
Thus, we find that:

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

### Final Answer
\[ \frac{\pi}{2}. \]

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