`(2)/(3!)+(4)/(5!)+(6)/(7!)+`..is equal to

Options

A. `e^(1//2)`
B. `e^(-1)`
C. e
D. `e^(-1//3)`

Answer

To solve the series \( S = \frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \ldots \), we start by identifying the general term of the series.

### Step-by-step Solution:

1. **Identify the General Term**:
The series can be expressed in terms of \( n \):
\[ S = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \]
This is because when \( n = 1 \), we have \( \frac{2 \cdot 1}{3!} \), when \( n = 2 \), we have \( \frac{2 \cdot 2}{5!} \), and so on.

2. **Rewrite the General Term**:
We can rewrite \( \frac{2n}{(2n+1)!} \) as:
\[ S = \sum_{n=1}^{\infty} \left( \frac{2n + 1 - 1}{(2n + 1)!} \right) \]
This simplifies to:
\[ S = \sum_{n=1}^{\infty} \left( \frac{2n + 1}{(2n + 1)!} - \frac{1}{(2n + 1)!} \right) \]

3. **Separate the Summation**:
Now we can separate the summation:
\[ S = \sum_{n=1}^{\infty} \frac{2n + 1}{(2n + 1)!} - \sum_{n=1}^{\infty} \frac{1}{(2n + 1)!} \]

4. **Evaluate the First Summation**:
The first summation can be simplified:
\[ \sum_{n=1}^{\infty} \frac{2n + 1}{(2n + 1)!} = \sum_{n=1}^{\infty} \frac{1}{(2n)!} = e^x \text{ evaluated at } x=1 \]
This is because the series for \( e^x \) includes all factorial terms.

5. **Evaluate the Second Summation**:
The second summation is:
\[ \sum_{n=1}^{\infty} \frac{1}{(2n + 1)!} = \frac{e - 1}{2} \]

6. **Combine the Results**:
Now we can combine the results:
\[ S = \sum_{n=1}^{\infty} \frac{1}{(2n)!} - \sum_{n=1}^{\infty} \frac{1}{(2n + 1)!} \]
This results in:
\[ S = e^{-1} \]

7. **Final Result**:
Therefore, the value of the series is:
\[ S = e - 1 \]