If `S_(n) = sum_(r=1)^(n) (r )/(1.3.5.7"……"(2r+1))`, then

Options

A. `S_(n) = 1/2 [1-(1)/(1.3.5"……."(2n+1))]`
B. `S_(oo) = 1/2`
C. `1/4 [1+(1)/(1.3.5"……"(2n-1))]`
D. `S_(oo) = 1/4`

Answer

To solve the problem, we need to analyze the given expression:

\[ S_n = \sum_{r=1}^{n} \frac{r}{1 \cdot 3 \cdot 5 \cdots (2r + 1)} \]

### Step 1: Understanding the Denominator
The denominator \(1 \cdot 3 \cdot 5 \cdots (2r + 1)\) can be expressed in terms of factorials. It represents the product of the first \(r\) odd numbers. This can be rewritten using the double factorial notation as:

\[ (2r + 1)!! = \frac{(2r + 1)!}{2^r \cdot r!} \]

### Step 2: Rewrite the Sum
Using the double factorial, we can rewrite \(S_n\) as:

\[ S_n = \sum_{r=1}^{n} \frac{r \cdot 2^r \cdot r!}{(2r + 1)!} \]

### Step 3: Simplifying the Expression
We can simplify the term \(\frac{r \cdot 2^r \cdot r!}{(2r + 1)!}\):

\[ \frac{r \cdot 2^r \cdot r!}{(2r + 1)!} = \frac{r \cdot 2^r}{(2r + 1)(2r)(2r - 1) \cdots (2)} = \frac{2^r}{(2r + 1)!!} \]

### Step 4: Evaluating the Limit as \(n \to \infty\)
To find \(S_\infty\), we take the limit as \(n\) approaches infinity:

\[ S_\infty = \lim_{n \to \infty} S_n \]

### Step 5: Analyzing the Options
We have the following options:
1. \(S_n = \frac{1}{2} \left(1 - \frac{1}{1 \cdot 3 \cdots (2n + 1)}\right)\)
2. \(S_\infty = \frac{1}{2}\)
3. \(S_n = \frac{1}{4} \left(1 + \frac{1}{1 \cdot 3 \cdots (2n - 1)}\right)\)
4. \(S_\infty = \frac{1}{4}\)

### Step 6: Checking the Options
- For option 1, substituting \(n = 1\):
\[ S_1 = \frac{1}{2} \left(1 - \frac{1}{3}\right) = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \]
- For option 2, \(S_\infty = \frac{1}{2}\) is valid.

- For option 3, substituting \(n = 1\):
\[ S_1 = \frac{1}{4} \left(1 + 1\right) = \frac{1}{4} \cdot 2 = \frac{1}{2} \]
- For option 4, \(S_\infty = \frac{1}{4}\) is not valid.

### Conclusion
The correct options are:
1. \(S_n = \frac{1}{2} \left(1 - \frac{1}{1 \cdot 3 \cdots (2n + 1)}\right)\)
2. \(S_\infty = \frac{1}{2}\)