If `P(u_i)) oo i` , where `i = 1, 2, 3, . . ., n`then `lim_(nrarrw) P(w)` is equal to

Options

A. `2/3`
B. `3/4`
C. `1/4`
D. 1

Answer

To solve the problem step by step, we need to find the limit of \( P(w) \) as \( n \) approaches infinity, given that \( P(u_i) \) is proportional to \( i \) for \( i = 1, 2, 3, \ldots, n \).

### Step-by-Step Solution:

1. **Understanding Proportionality**:
Since \( P(u_i) \) is proportional to \( i \), we can express this as:
\[ P(u_i) = k \cdot i \]
where \( k \) is a constant.

2. **Summation of Probabilities**:
The sum of all probabilities must equal 1:
\[ \sum_{i=1}^{n} P(u_i) = 1 \]
Substituting \( P(u_i) \):
\[ \sum_{i=1}^{n} k \cdot i = 1 \]
This simplifies to:
\[ k \cdot \sum_{i=1}^{n} i = 1 \]

3. **Using the Formula for the Sum of First n Natural Numbers**:
The sum of the first \( n \) natural numbers is given by:
\[ \sum_{i=1}^{n} i = \frac{n(n + 1)}{2} \]
Therefore, we have:
\[ k \cdot \frac{n(n + 1)}{2} = 1 \]
From this, we can solve for \( k \):
\[ k = \frac{2}{n(n + 1)} \]

4. **Finding \( P(w) \)**:
We need to find \( P(w) \):
\[ P(w) = k \cdot w = \frac{2w}{n(n + 1)} \]

5. **Taking the Limit as \( n \to \infty \)**:
Now we take the limit of \( P(w) \) as \( n \) approaches infinity:
\[ \lim_{n \to \infty} P(w) = \lim_{n \to \infty} \frac{2w}{n(n + 1)} \]
As \( n \) approaches infinity, \( n(n + 1) \) also approaches infinity, thus:
\[ \lim_{n \to \infty} P(w) = 0 \]

### Final Answer:
\[ \lim_{n \to \infty} P(w) = 0 \]