`1+ 1/(2!)+2/(3!)+4/(4!)+ .... oo = `

Options

A. `e^2`
B. `e^(2) +1`
C. `e^(2) -1`
D. `1/4 (e^(2) +1)`

Answer

To solve the series \( S = 1 + \frac{1}{2!} + \frac{2}{3!} + \frac{4}{4!} + \ldots \) up to infinity, we can use the properties of the exponential function.

### Step-by-step Solution:

1. **Identify the Series**:
The series can be expressed as:
\[ S = \sum_{n=0}^{\infty} \frac{a_n}{n!} \]
where \( a_0 = 1, a_1 = 1, a_2 = 2, a_3 = 4, \ldots \)

2. **Recognize the Pattern**:
The coefficients \( a_n \) seem to follow a pattern. We can observe that:
\[ a_n = 2^{n-1} \text{ for } n \geq 1 \]
Therefore, we can rewrite the series as:
\[ S = 1 + \sum_{n=1}^{\infty} \frac{2^{n-1}}{n!} \]

3. **Factor Out Constants**:
We can factor out \( 2 \) from the summation:
\[ S = 1 + 2 \sum_{n=1}^{\infty} \frac{2^{n-2}}{(n-1)!} \]
By changing the index of summation (let \( m = n-1 \)), we have:
\[ S = 1 + 2 \sum_{m=0}^{\infty} \frac{2^{m}}{m!} \]

4. **Recognize the Exponential Function**:
The series \( \sum_{m=0}^{\infty} \frac{2^{m}}{m!} \) is the Taylor series expansion for \( e^x \) evaluated at \( x = 2 \):
\[ \sum_{m=0}^{\infty} \frac{2^{m}}{m!} = e^2 \]

5. **Substitute Back**:
Now substituting back into the expression for \( S \):
\[ S = 1 + 2e^2 \]

6. **Final Expression**:
Thus, the series converges to:
\[ S = 1 + 2e^2 \]

### Conclusion:
The final result of the series is:
\[ S = 1 + 2e^2 \]