The value of `lim_(n to oo)[(n)/(n^(2))+(n)/(n^(2)+1^(2))+(n)/(n^(2)+2^(2))+...+(n)/(n^(2)+(n-1)^(2))]` is :

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `pi`
D. `-(pi)/(4)`

Answer

To solve the limit

\[ \lim_{n \to \infty} \left[ \frac{n}{n^2} + \frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + \ldots + \frac{n}{n^2 + (n-1)^2} \right], \]

we can break down the solution step by step.

### Step 1: Rewrite the expression
The limit can be rewritten as:

\[ \lim_{n \to \infty} \sum_{r=0}^{n-1} \frac{n}{n^2 + r^2}. \]

### Step 2: Factor out \( n \)
We can factor \( n \) out of the denominator:

\[ \frac{n}{n^2 + r^2} = \frac{n}{n^2(1 + \frac{r^2}{n^2})} = \frac{1}{n(1 + \frac{r^2}{n^2})}. \]

Thus, the sum becomes:

\[ \lim_{n \to \infty} \sum_{r=0}^{n-1} \frac{1}{n(1 + \frac{r^2}{n^2})}. \]

### Step 3: Change the sum to an integral
As \( n \to \infty \), we can interpret the sum as a Riemann sum. Let \( x = \frac{r}{n} \), then \( r = nx \) and \( \frac{dr}{n} = dx \). The limits for \( r \) from \( 0 \) to \( n-1 \) correspond to \( x \) from \( 0 \) to \( 1 - \frac{1}{n} \), which approaches \( 0 \) to \( 1 \) as \( n \to \infty \).

The sum can be approximated by the integral:

\[ \int_0^1 \frac{1}{1 + x^2} \, dx. \]

### Step 4: Evaluate the integral
The integral

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

is a standard integral, which evaluates to:

\[ \left[ \tan^{-1}(x) \right]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

### Step 5: Conclusion
Thus, the value of the limit is:

\[ \lim_{n \to \infty} \left[ \frac{n}{n^2} + \frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + \ldots + \frac{n}{n^2 + (n-1)^2} \right] = \frac{\pi}{4}. \]