Let `a_1=1`, `a_n=n(a_(n-1)+1)` for `n=2,3,...` where `P_n=(1+1/a_1)(1+1/a_2)(1+1/a_3)....(1+1/a_n)` then `lim_(nrarroo)P_n=`

Options

A. `e`
B. `e/2`
C. `2e`
D. `3e`

Answer

To solve the problem, we need to find the limit of \( P_n \) as \( n \) approaches infinity, where

\[ P_n = \left(1 + \frac{1}{a_1}\right) \left(1 + \frac{1}{a_2}\right) \left(1 + \frac{1}{a_3}\right) \cdots \left(1 + \frac{1}{a_n}\right) \]

with the recursive definition of \( a_n \) given by:

\[ a_1 = 1, \quad a_n = n(a_{n-1} + 1) \quad \text{for } n = 2, 3, \ldots \]

### Step 1: Calculate the first few terms of \( a_n \)

1. **Calculate \( a_2 \)**:
\[ a_2 = 2(a_1 + 1) = 2(1 + 1) = 2 \times 2 = 4 \]

2. **Calculate \( a_3 \)**:
\[ a_3 = 3(a_2 + 1) = 3(4 + 1) = 3 \times 5 = 15 \]

3. **Calculate \( a_4 \)**:
\[ a_4 = 4(a_3 + 1) = 4(15 + 1) = 4 \times 16 = 64 \]

From these calculations, we can see a pattern forming.

### Step 2: Generalize \( a_n \)

We can see that:
- \( a_1 = 1 \)
- \( a_2 = 4 \)
- \( a_3 = 15 \)
- \( a_4 = 64 \)

This suggests that \( a_n \) grows quite rapidly. We can express \( a_n \) in terms of factorials:

\[ a_n = n! \cdot \text{(some function of } n\text{)} \]

### Step 3: Analyze \( P_n \)

Now we can express \( P_n \):

\[ P_n = \prod_{k=1}^{n} \left(1 + \frac{1}{a_k}\right) \]

Using the values we calculated:
\[ P_n = \left(1 + \frac{1}{1}\right) \left(1 + \frac{1}{4}\right) \left(1 + \frac{1}{15}\right) \left(1 + \frac{1}{64}\right) \cdots \]

### Step 4: Simplify \( P_n \)

As \( n \) becomes very large, \( a_n \) becomes very large, and thus \( \frac{1}{a_k} \) becomes very small. We can approximate:

\[ P_n \approx \prod_{k=1}^{n} \left(1 + \frac{1}{a_k}\right) \approx \prod_{k=1}^{n} e^{\frac{1}{a_k}} = e^{\sum_{k=1}^{n} \frac{1}{a_k}} \]

### Step 5: Find the limit

To find the limit of \( P_n \) as \( n \) approaches infinity, we need to evaluate \( \sum_{k=1}^{n} \frac{1}{a_k} \). As \( a_k \) grows rapidly, this sum converges.

### Conclusion

Thus, we can conclude that:

\[ \lim_{n \to \infty} P_n = e^{\lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{a_k}} = e \]

So the final answer is:

\[ \lim_{n \to \infty} P_n = e \]