Evaluate `lim_(n->oo)1/nsum_(r=n+1)^(2n)log_e(1+r/n)`

Options

A. `log((27)/(4e))`
B. `log((27)/(e^(2)))`
C. `log((4)/(e ))`
D. None of these

Answer

To evaluate the limit

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=n+1}^{2n} \log\left(1 + \frac{r}{n}\right), \]

we can follow these steps:

### Step 1: Rewrite the summation
We start by rewriting the summation term:

\[ \log\left(1 + \frac{r}{n}\right). \]

### Step 2: Change the variable
Let \( x = \frac{r}{n} \). Then, when \( r = n + 1 \), \( x \to 1 + \frac{1}{n} \) and when \( r = 2n \), \( x \to 2 \). The increment \( dr = n \, dx \).

### Step 3: Convert the summation to an integral
The limit can be expressed as:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=n+1}^{2n} \log\left(1 + \frac{r}{n}\right) = \lim_{n \to \infty} \sum_{r=n+1}^{2n} \frac{1}{n} \log\left(1 + \frac{r}{n}\right) \approx \int_{1}^{2} \log(1 + x) \, dx. \]

### Step 4: Evaluate the integral
Now we need to evaluate the integral:

\[ \int_{1}^{2} \log(1 + x) \, dx. \]

Using integration by parts, let \( u = \log(1 + x) \) and \( dv = dx \). Then \( du = \frac{1}{1+x} \, dx \) and \( v = x \).

Using integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we have:

\[ \int \log(1 + x) \, dx = x \log(1 + x) - \int \frac{x}{1 + x} \, dx. \]

### Step 5: Simplify the remaining integral
The integral \( \int \frac{x}{1 + x} \, dx \) can be simplified as follows:

\[ \int \frac{x}{1+x} \, dx = \int \left(1 - \frac{1}{1+x}\right) \, dx = x - \log(1+x). \]

### Step 6: Combine results
Thus, we have:

\[ \int \log(1 + x) \, dx = x \log(1 + x) - (x - \log(1 + x)) = (x + 1) \log(1 + x) - x. \]

Now evaluate from 1 to 2:

\[ \left[(x + 1) \log(1 + x) - x\right]_{1}^{2}. \]

Calculating at the bounds:

1. At \( x = 2 \):

\[ (2 + 1) \log(3) - 2 = 3 \log(3) - 2. \]

2. At \( x = 1 \):

\[ (1 + 1) \log(2) - 1 = 2 \log(2) - 1. \]

### Step 7: Final calculation
Thus, we have:

\[ \int_{1}^{2} \log(1 + x) \, dx = \left[3 \log(3) - 2\right] - \left[2 \log(2) - 1\right] = 3 \log(3) - 2 - 2 \log(2) + 1 = 3 \log(3) - 2 \log(2) - 1. \]

### Step 8: Simplify the expression
This can be expressed as:

\[ 3 \log(3) - 2 \log(2) - 1 = \log(3^3) - \log(2^2) - \log(e) = \log\left(\frac{27}{4e}\right). \]

### Final Result
Thus, the limit evaluates to:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=n+1}^{2n} \log\left(1 + \frac{r}{n}\right) = \log\left(\frac{27}{4e}\right). \]