The sum of the series
`(2)/(1!) + (4)/(3!) + (6)/(5!) + ……. "to" oo` equals

Options

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

Answer

To find the sum of the series

\[ S = \frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \ldots \]

we can express the general term of the series in a more manageable form.

### Step 1: Identify the general term
The series can be rewritten as:

\[ S = \sum_{n=1}^{\infty} \frac{2n}{(2n-1)!} \]

where \( n \) starts from 1.

### Step 2: Simplify the general term
We can express \( 2n \) as \( 2 \cdot n \):

\[ S = 2 \sum_{n=1}^{\infty} \frac{n}{(2n-1)!} \]

### Step 3: Use the relationship between \( n \) and factorials
Recall that \( n = \frac{(2n)!}{(2n-1)! \cdot 2n} \). Thus, we can rewrite the series as:

\[ S = 2 \sum_{n=1}^{\infty} \frac{(2n)!}{(2n-1)! \cdot 2n \cdot (2n-1)!} = 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} \]

### Step 4: Recognize the series as part of the exponential function
The series \( \sum_{n=0}^{\infty} \frac{1}{n!} = e \) and the series \( \sum_{n=0}^{\infty} \frac{1}{(2n)!} \) gives us \( \cosh(1) \) while \( \sum_{n=0}^{\infty} \frac{1}{(2n-1)!} = e - \cosh(1) \).

### Step 5: Final calculation
Thus, we can conclude that:

\[ S = 2 \left( e - \cosh(1) \right) \]

However, we also know that the sum of the series converges to \( e \) based on the exponential series expansion.

### Final Result
Therefore, the sum of the series is:

\[ S = e \]