The value of ` lim_(nto oo)(1)/(2) sum_(r-1)^(n) ((r)/(n+r))` is equal to

Options

A. `1 - log 2`
B. `log 4-1`
C. ` log 2`
D. None of these

Answer

To solve the limit problem, we need to evaluate:

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

### Step 1: Rewrite the expression
We start by rewriting the sum in a more manageable form:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{r}{n+r} = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{r/n}{1 + r/n} \]

### Step 2: Change of variable
Let \( x = \frac{r}{n} \). As \( n \to \infty \), \( r \) varies from \( 1/n \) to \( 1 \). The sum can be approximated by a Riemann sum:

\[ \sum_{r=1}^{n} \frac{r/n}{1 + r/n} \cdot \frac{1}{n} \approx \int_{0}^{1} \frac{x}{1+x} \, dx \]

### Step 3: Set up the integral
Thus, we rewrite our limit as an integral:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{r/n}{1 + r/n} \approx \int_{0}^{1} \frac{x}{1+x} \, dx \]

### Step 4: Evaluate the integral
Now we compute the integral:

\[ \int_{0}^{1} \frac{x}{1+x} \, dx \]

We can simplify the integrand:

\[ \frac{x}{1+x} = 1 - \frac{1}{1+x} \]

Thus, the integral becomes:

\[ \int_{0}^{1} \left( 1 - \frac{1}{1+x} \right) \, dx = \int_{0}^{1} 1 \, dx - \int_{0}^{1} \frac{1}{1+x} \, dx \]

### Step 5: Calculate the integrals
Calculating the first integral:

\[ \int_{0}^{1} 1 \, dx = 1 \]

Calculating the second integral:

\[ \int_{0}^{1} \frac{1}{1+x} \, dx = \left[ \ln(1+x) \right]_{0}^{1} = \ln(2) - \ln(1) = \ln(2) \]

### Step 6: Combine results
Putting it all together:

\[ \int_{0}^{1} \frac{x}{1+x} \, dx = 1 - \ln(2) \]

### Final Result
Thus, the value of the limit is:

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