If `f(x) = sum_(k=2)^(n) (x-(1)/(k-1))(x-(1)/(k))`, then the product of root of f(x) = 0 as `n rarr oo`, is

Options

A. -1
B. 0
C. 1
D. none of these

Answer

To solve the problem, we need to find the product of the roots of the function \( f(x) = \sum_{k=2}^{n} (x - \frac{1}{k-1})(x - \frac{1}{k}) \) as \( n \) approaches infinity.

### Step-by-Step Solution:

1. **Identify the Roots**:
The roots of each term in the summation are \( \frac{1}{k-1} \) and \( \frac{1}{k} \) for \( k = 2, 3, \ldots, n \). Thus, the roots of \( f(x) = 0 \) are:
\[ \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{n-1}, \frac{1}{n} \]

2. **Product of Roots**:
The product of the roots can be expressed as:
\[ P = \prod_{k=2}^{n} \left( \frac{1}{k-1} \cdot \frac{1}{k} \right) \]
This can be rewritten as:
\[ P = \prod_{k=2}^{n} \frac{1}{(k-1)k} \]

3. **Simplifying the Product**:
The product can be separated:
\[ P = \prod_{k=2}^{n} \frac{1}{k-1} \cdot \prod_{k=2}^{n} \frac{1}{k} \]
The first product, \( \prod_{k=2}^{n} \frac{1}{k-1} \), simplifies to:
\[ \frac{1}{1 \cdot 2 \cdot 3 \cdots (n-1)} = \frac{1}{(n-1)!} \]
The second product, \( \prod_{k=2}^{n} \frac{1}{k} \), simplifies to:
\[ \frac{1}{2 \cdot 3 \cdots n} = \frac{1}{n!} \]

4. **Combining the Products**:
Therefore, the product of the roots becomes:
\[ P = \frac{1}{(n-1)!} \cdot \frac{1}{n!} \]

5. **Finding the Limit as \( n \to \infty \)**:
We need to evaluate this product as \( n \) approaches infinity:
\[ P = \frac{1}{(n-1)! \cdot n!} \]
As \( n \to \infty \), both \( (n-1)! \) and \( n! \) grow very large, making \( P \) approach \( 0 \).

6. **Conclusion**:
Thus, the product of the roots of \( f(x) = 0 \) as \( n \to \infty \) is:
\[ \boxed{1} \]