If `a,b in R^(+)` then find `Lim_(nrarroo) sum_(k=1)^(n) ( n)/((k+an)(k+bn))` is equal to

Options

A. `(1)/(a-b) ln'(b(b+1))/(a(a+1))` if `a ne b`
B. `(1)/(a-b) ln'(a(b+1))/(b(a+1))` if `a ne b`
C. non existent if `a = b`
D. `(1)/(a(1+a))` if `a ne b`

Answer

To solve the limit \( \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n}{(k + an)(k + bn)} \), we will follow these steps:

### Step 1: Rewrite the Summation
We start by rewriting the expression inside the summation:
\[ \frac{n}{(k + an)(k + bn)} = \frac{n}{n^2 \left(\frac{k}{n} + a\right)\left(\frac{k}{n} + b\right)} = \frac{1}{n \left(\frac{k}{n} + a\right)\left(\frac{k}{n} + b\right)} \]

### Step 2: Change of Variable
Let \( x = \frac{k}{n} \). Then, as \( k \) goes from \( 1 \) to \( n \), \( x \) goes from \( \frac{1}{n} \) to \( 1 \). The increment \( \Delta x = \frac{1}{n} \).

Thus, we can rewrite the summation as:
\[ \sum_{k=1}^{n} \frac{1}{n \left(\frac{k}{n} + a\right)\left(\frac{k}{n} + b\right)} \approx \sum_{k=1}^{n} \frac{1}{n} \cdot \frac{1}{(x + a)(x + b)} \]

### Step 3: Convert to Integral
As \( n \to \infty \), the sum approaches the integral:
\[ \int_{0}^{1} \frac{1}{(x + a)(x + b)} \, dx \]

### Step 4: Solve the Integral
To solve the integral, we can use partial fraction decomposition:
\[ \frac{1}{(x + a)(x + b)} = \frac{1}{b - a} \left( \frac{1}{x + a} - \frac{1}{x + b} \right) \]

### Step 5: Integrate
Now we can integrate:
\[ \int_{0}^{1} \left( \frac{1}{b - a} \left( \frac{1}{x + a} - \frac{1}{x + b} \right) \right) dx = \frac{1}{b - a} \left( \left[ \ln(x + a) \right]_{0}^{1} - \left[ \ln(x + b) \right]_{0}^{1} \right) \]

Calculating the limits:
\[ = \frac{1}{b - a} \left( \ln(1 + a) - \ln(a) - \ln(1 + b) + \ln(b) \right) \]
\[ = \frac{1}{b - a} \left( \ln\left(\frac{1 + a}{a}\right) - \ln\left(\frac{1 + b}{b}\right) \right) \]
\[ = \frac{1}{b - a} \ln\left(\frac{(1 + a)b}{(1 + b)a}\right) \]

### Final Answer
Thus, the limit is:
\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n}{(k + an)(k + bn)} = \frac{1}{b - a} \ln\left(\frac{(1 + a)b}{(1 + b)a}\right) \]