Sum the series `(2)/(3!) + (4)/(5!) + (6)/(7!) + (8)/(9!) + ...`

Options

A. e
B. 2e
C. `e^(-1)`
D. None

Answer

To sum the series \( S = \frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \frac{8}{9!} + \ldots \), we can follow these steps:

### Step 1: Rewrite the terms of the series
We can express each term in the series in a different form. Notice that:
- \( 2 = 3 - 1 \)
- \( 4 = 5 - 1 \)
- \( 6 = 7 - 1 \)
- \( 8 = 9 - 1 \)

Thus, we can rewrite the series as:
\[ S = \left( \frac{3 - 1}{3!} \right) + \left( \frac{5 - 1}{5!} \right) + \left( \frac{7 - 1}{7!} \right) + \left( \frac{9 - 1}{9!} \right) + \ldots \]

### Step 2: Separate the series into two parts
Now, we can separate the series into two distinct sums:
\[ S = \left( \frac{3}{3!} + \frac{5}{5!} + \frac{7}{7!} + \ldots \right) - \left( \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \ldots \right) \]

Let:
\[ S_1 = \frac{3}{3!} + \frac{5}{5!} + \frac{7}{7!} + \ldots \]
\[ S_2 = \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \ldots \]

Thus, we have:
\[ S = S_1 - S_2 \]

### Step 3: Simplify \( S_1 \)
We can express \( S_1 \) as:
\[ S_1 = \sum_{n=1}^{\infty} \frac{2n + 1}{(2n + 1)!} \]
This can be rewritten as:
\[ S_1 = \sum_{n=1}^{\infty} \frac{2n}{(2n + 1)!} + \sum_{n=1}^{\infty} \frac{1}{(2n + 1)!} \]
The first part can be simplified using the identity \( \frac{2n}{(2n + 1)!} = \frac{2}{(2n)!} \):
\[ S_1 = 2 \sum_{n=1}^{\infty} \frac{1}{(2n)!} + \sum_{n=1}^{\infty} \frac{1}{(2n + 1)!} \]

### Step 4: Evaluate \( S_2 \)
The series \( S_2 \) can be recognized as half of the series for \( e^x \):
\[ S_2 = \sum_{n=1}^{\infty} \frac{1}{(2n)!} = \frac{1}{2} \left( e - 1 \right) \]

### Step 5: Combine the results
Now we can combine the results of \( S_1 \) and \( S_2 \):
\[ S = S_1 - S_2 \]
Using the known expansions for \( e^x \), we find:
\[ S_1 = e - 1 - S_2 \]
Substituting \( S_2 \):
\[ S = (e - 1) - \frac{1}{2}(e - 1) = \frac{1}{2}(e - 1) \]

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