The sum of the series `(2)/(3!)+(4)/(5!)+(6)/(7!)+. . . .` is

Options

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

Answer

To find the sum of the series \( S = \frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \ldots \), we can follow these steps:

### Step 1: Rewrite the Series
The series can be expressed in a more general form. We can rewrite the terms as follows:
\[ S = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \]

### Step 2: Separate the Terms
We can separate the term \( \frac{2n}{(2n+1)!} \) into two parts:
\[ \frac{2n}{(2n+1)!} = \frac{2n + 1 - 1}{(2n+1)!} = \frac{2n + 1}{(2n+1)!} - \frac{1}{(2n+1)!} \]
Thus, we can rewrite \( S \) as:
\[ S = \sum_{n=1}^{\infty} \left( \frac{2n + 1}{(2n + 1)!} - \frac{1}{(2n + 1)!} \right) \]

### Step 3: Simplify the Series
This can be simplified into two separate series:
\[ S = \sum_{n=1}^{\infty} \frac{2n + 1}{(2n + 1)!} - \sum_{n=1}^{\infty} \frac{1}{(2n + 1)!} \]

### Step 4: Evaluate the First Series
The first series can be evaluated using the known series expansion for \( e^x \):
\[ \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \]
For our case, we can relate it to the series for \( e^x \):
\[ \sum_{n=0}^{\infty} \frac{1}{(2n)!} = \cosh(1) \quad \text{and} \quad \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} = \sinh(1) \]
Thus, we have:
\[ \sum_{n=1}^{\infty} \frac{2n + 1}{(2n + 1)!} = \sinh(1) \]

### Step 5: Evaluate the Second Series
The second series can be evaluated as:
\[ \sum_{n=1}^{\infty} \frac{1}{(2n + 1)!} = \sinh(1) - 1 \]

### Step 6: Combine the Results
Now we can combine the results:
\[ S = \sinh(1) - (\sinh(1) - 1) = 1 \]

### Final Result
Thus, the sum of the series is:
\[ \boxed{1} \]