The value of `I=lim_(nrarroo)Sigma_(r=1)^(n)(r)/(n^(2)+n+r)` is equal to

Options

A. `(1)/(3)`
B. `(1)/(2)`
C. `(pi)/(2)`
D. `(pi)/(3)`

Answer

To solve the limit \( I = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r}{n^2 + n + r} \), we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression inside the limit:
\[ I = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r}{n^2 + n + r} \]
Notice that as \( n \) becomes very large, \( n^2 + n + r \) can be approximated by \( n^2 \) since \( r \) is much smaller than \( n^2 \).

### Step 2: Simplify the fraction
We can factor \( n^2 \) out of the denominator:
\[ I = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r}{n^2(1 + \frac{n}{n^2} + \frac{r}{n^2})} = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r/n^2}{1 + \frac{1}{n} + \frac{r}{n^2}} \]
This simplifies to:
\[ I = \lim_{n \to \infty} \frac{1}{n^2} \sum_{r=1}^{n} \frac{r}{1 + \frac{1}{n} + \frac{r}{n^2}} \]

### Step 3: Evaluate the sum
The sum \( \sum_{r=1}^{n} r \) is known to be:
\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]
Thus, we can rewrite \( I \):
\[ I = \lim_{n \to \infty} \frac{1}{n^2} \cdot \frac{n(n + 1)}{2} \cdot \frac{1}{1 + \frac{1}{n} + \frac{n(n+1)}{2n^2}} \]

### Step 4: Simplify further
As \( n \to \infty \), \( \frac{1}{n} \to 0 \) and \( \frac{n(n+1)}{2n^2} \to \frac{1}{2} \):
\[ I = \lim_{n \to \infty} \frac{1}{n^2} \cdot \frac{n(n + 1)}{2} \cdot \frac{1}{1 + 0 + \frac{1}{2}} = \lim_{n \to \infty} \frac{n(n + 1)}{4n^2} = \lim_{n \to \infty} \frac{n^2 + n}{4n^2} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{4} \]

### Step 5: Take the limit
Taking the limit as \( n \to \infty \):
\[ I = \frac{1 + 0}{4} = \frac{1}{4} \]

### Conclusion
Thus, the value of \( I \) is:
\[ \boxed{\frac{1}{4}} \]