Let `S_(n)=1+2+3+...+n " and " P_(n)=(S_(2))/(S_(2)-1).(S_(3))/(S_(3)-1).(S_(4))/(S_(4)-1)...(S_(n))/(S_(n)-1)`, where `n inN,(nge2) Then ``underset(ntooo)limP_(n)`=__________.

Options

A. `(1)/(4)`
B. `(1)/(24)`
C. 3
D. `(1)/(8)`

Explanation

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

\[ S_n = 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \]

and

\[ P_n = \frac{S_2}{S_2 - 1} \cdot \frac{S_3}{S_3 - 1} \cdot \frac{S_4}{S_4 - 1} \cdots \frac{S_n}{S_n - 1}. \]

### Step 1: Calculate \( S_k \) and \( S_k - 1 \)

For each \( k \):

\[ S_k = \frac{k(k+1)}{2} \]

Thus,

\[ S_k - 1 = \frac{k(k+1)}{2} - 1 = \frac{k(k+1) - 2}{2} = \frac{k^2 + k - 2}{2} = \frac{(k-1)(k+2)}{2}. \]

### Step 2: Calculate \( \frac{S_k}{S_k - 1} \)

Now we can find \( \frac{S_k}{S_k - 1} \):

\[ \frac{S_k}{S_k - 1} = \frac{\frac{k(k+1)}{2}}{\frac{(k-1)(k+2)}{2}} = \frac{k(k+1)}{(k-1)(k+2)}. \]

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

Now we can express \( P_n \):

\[ P_n = \prod_{k=2}^{n} \frac{k(k+1)}{(k-1)(k+2)}. \]

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

We can simplify \( P_n \):

\[ P_n = \frac{2 \cdot 3}{1 \cdot 4} \cdot \frac{3 \cdot 4}{2 \cdot 5} \cdot \frac{4 \cdot 5}{3 \cdot 6} \cdots \frac{n(n+1)}{(n-1)(n+2)}. \]

Notice that this is a telescoping product. Most terms will cancel out:

\[ P_n = \frac{2}{1} \cdot \frac{n(n+1)}{(n-1)(n+2)}. \]

### Step 5: Take the limit as \( n \to \infty \)

Now we need to evaluate the limit:

\[ \lim_{n \to \infty} P_n = \lim_{n \to \infty} \frac{2 \cdot n(n+1)}{(n-1)(n+2)}. \]

As \( n \) approaches infinity, we can simplify:

\[ \frac{n(n+1)}{(n-1)(n+2)} \approx \frac{n^2}{n^2} = 1. \]

Thus,

\[ \lim_{n \to \infty} P_n = 2 \cdot 1 = 2. \]

### Conclusion

The limit \( \lim_{n \to \infty} P_n = 2 \).

### Final Answer

\[ \text{The limit } \lim_{n \to \infty} P_n = 2. \]

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