`1/(3!) +2/(5!) + (3)/(7!) + .....oo` =

Options

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

Answer

To solve the series \( S = \frac{1}{3!} + \frac{2}{5!} + \frac{3}{7!} + \ldots \) up to infinity, we will follow these steps:

### Step 1: Rewrite the Series
We start by rewriting the series in a more manageable form:
\[ S = \sum_{n=1}^{\infty} \frac{n}{(2n + 1)!} \]
This represents the series where \( n \) takes values starting from 1 and goes to infinity.

### Step 2: Factor Out a Constant
To simplify our calculations, we can factor out a constant. We multiply and divide the series by 2:
\[ S = \frac{1}{2} \sum_{n=1}^{\infty} \frac{2n}{(2n + 1)!} \]

### Step 3: Rewrite Each Term
Next, we rewrite each term in the series:
\[ \frac{2n}{(2n + 1)!} = \frac{(2n + 1) - 1}{(2n + 1)!} = \frac{2n + 1}{(2n + 1)!} - \frac{1}{(2n + 1)!} \]
This allows us to break the series into two separate sums:
\[ S = \frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{1}{(2n)!} - \sum_{n=1}^{\infty} \frac{1}{(2n + 1)!} \right) \]

### Step 4: Recognize the Series
The first series \( \sum_{n=0}^{\infty} \frac{1}{(2n)!} \) is the Taylor series expansion for \( \cosh(1) \), and the second series \( \sum_{n=0}^{\infty} \frac{1}{(2n + 1)!} \) is the Taylor series expansion for \( \sinh(1) \):
\[ \sum_{n=0}^{\infty} \frac{1}{(2n)!} = \cosh(1), \quad \sum_{n=0}^{\infty} \frac{1}{(2n + 1)!} = \sinh(1) \]

### Step 5: Combine the Results
Thus, we can express our series \( S \) as:
\[ S = \frac{1}{2} \left( \cosh(1) - \sinh(1) \right) \]

### Step 6: Simplify Further
Using the definitions of hyperbolic functions, we know that:
\[ \cosh(1) - \sinh(1) = e^{-1} \]
Therefore, we have:
\[ S = \frac{1}{2} e^{-1} \]

### Final Result
Thus, the value of the series is:
\[ S = \frac{1}{2e} \]