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 use the concept of the Squeeze Theorem. Here’s a step-by-step solution:

### Step 1: Analyze the expression
We start with the expression inside the limit:
\[ I = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r}{n^2 + n + r} \]
As \( n \) becomes very large, we can observe that \( n^2 + n + r \) behaves similarly to \( n^2 \) since \( r \) is much smaller than \( n^2 \) for large \( n \).

### Step 2: Establish bounds for the summation
We can establish lower and upper bounds for the term \( \frac{r}{n^2 + n + r} \):
- **Lower Bound**:
\[ \frac{r}{n^2 + n + r} \geq \frac{r}{n^2 + n + n} = \frac{r}{n^2 + 2n} \]
- **Upper Bound**:
\[ \frac{r}{n^2 + n + r} \leq \frac{r}{n^2} \]

### Step 3: Sum the bounds
Now we can sum these bounds from \( r = 1 \) to \( n \):
1. **Lower Bound**:
\[ \sum_{r=1}^{n} \frac{r}{n^2 + 2n} = \frac{1}{n^2 + 2n} \sum_{r=1}^{n} r = \frac{1}{n^2 + 2n} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{2(n^2 + 2n)} \]
2. **Upper Bound**:
\[ \sum_{r=1}^{n} \frac{r}{n^2} = \frac{1}{n^2} \sum_{r=1}^{n} r = \frac{1}{n^2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{2n^2} \]

### Step 4: Simplify the bounds
Now we simplify both bounds:
- **Lower Bound**:
\[ \frac{n(n+1)}{2(n^2 + 2n)} = \frac{n(n+1)}{2n^2(1 + \frac{2}{n})} \approx \frac{1}{2} \text{ as } n \to \infty \]
- **Upper Bound**:
\[ \frac{n(n+1)}{2n^2} = \frac{1}{2} \left(1 + \frac{1}{n}\right) \approx \frac{1}{2} \text{ as } n \to \infty \]

### Step 5: Apply the Squeeze Theorem
Since both the lower and upper bounds converge to \( \frac{1}{2} \) as \( n \to \infty \), by the Squeeze Theorem, we conclude that:
\[ I = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r}{n^2 + n + r} = \frac{1}{2} \]

### Final Answer
Thus, the value of \( I \) is:
\[ \boxed{\frac{1}{2}} \]