Let a sequence whose `n^(th)` term is `{a_(n)}` be defined as
`a_(1) = 1/2` and `(n-1)a_(n-1) = (n+1)a_(n)` for `n ge 2` then `Lim_(n rarroo) S_(n)` equals

Options

A. `3/4`
B. `5/4`
C. `3/2`
D. `1`

Answer

To solve the problem, we need to find the limit of the sequence defined by the recurrence relation and the initial term provided. Let's break it down step by step.

### Step 1: Understand the recurrence relation
The nth term of the sequence is defined by:
- \( a_1 = \frac{1}{2} \)
- For \( n \geq 2 \), the relation is given by:
\[ (n-1) a_{n-1} = (n+1) a_n \]
From this, we can express \( a_n \) in terms of \( a_{n-1} \):
\[ a_n = \frac{(n-1)}{(n+1)} a_{n-1} \]

### Step 2: Calculate the first few terms
Now we can compute the first few terms of the sequence using the recurrence relation.

- **For \( n = 2 \)**:
\[ a_2 = \frac{(2-1)}{(2+1)} a_1 = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} \]

- **For \( n = 3 \)**:
\[ a_3 = \frac{(3-1)}{(3+1)} a_2 = \frac{2}{4} \cdot \frac{1}{6} = \frac{1}{12} \]

- **For \( n = 4 \)**:
\[ a_4 = \frac{(4-1)}{(4+1)} a_3 = \frac{3}{5} \cdot \frac{1}{12} = \frac{1}{20} \]

### Step 3: General term of the sequence
We observe a pattern in the terms calculated. We can generalize:
\[ a_n = \frac{1}{n^2 + n} \]
This can be verified by induction or by observing the pattern from the computed terms.

### Step 4: Find the sum \( S_n \)
The sum \( S_n \) of the first \( n \) terms is given by:
\[ S_n = \sum_{r=1}^{n} a_r = \sum_{r=1}^{n} \frac{1}{r^2 + r} \]
We can simplify \( \frac{1}{r^2 + r} \):
\[ \frac{1}{r^2 + r} = \frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1} \]
Thus, the sum becomes:
\[ S_n = \sum_{r=1}^{n} \left( \frac{1}{r} - \frac{1}{r+1} \right) \]

### Step 5: Evaluate the telescoping series
This is a telescoping series:
\[ S_n = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \]
Most terms cancel out, leaving us with:
\[ S_n = 1 - \frac{1}{n+1} \]

### Step 6: Find the limit as \( n \to \infty \)
Now we take the limit as \( n \) approaches infinity:
\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1 - 0 = 1 \]

### Final Answer
Thus, the limit of the sum \( S_n \) as \( n \) approaches infinity is:
\[ \boxed{1} \]