` sum _(r =1) ^(oo) (1+a+a^(2)+... +a^(r-1))/(r!)` is equal to

Options

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

Explanation

To solve the problem, we need to evaluate the infinite series given by:

\[ S = \sum_{r=1}^{\infty} \frac{1 + a + a^2 + \ldots + a^{r-1}}{r!} \]

### Step 1: Recognize the series in the numerator
The series \(1 + a + a^2 + \ldots + a^{r-1}\) is a geometric series with the first term \(1\) and common ratio \(a\). The sum of the first \(r\) terms of a geometric series can be expressed as:

\[ \text{Sum} = \frac{1 - a^r}{1 - a} \quad \text{(for } a \neq 1\text{)} \]

### Step 2: Substitute the sum of the geometric series into the original series
Substituting this sum into our series, we get:

\[ S = \sum_{r=1}^{\infty} \frac{\frac{1 - a^r}{1 - a}}{r!} = \frac{1}{1 - a} \sum_{r=1}^{\infty} \frac{1 - a^r}{r!} \]

### Step 3: Split the series
We can split the series into two parts:

\[ S = \frac{1}{1 - a} \left( \sum_{r=1}^{\infty} \frac{1}{r!} - \sum_{r=1}^{\infty} \frac{a^r}{r!} \right) \]

### Step 4: Evaluate the series
The first series, \(\sum_{r=1}^{\infty} \frac{1}{r!}\), is the Taylor series expansion for \(e\) (the base of natural logarithms):

\[ \sum_{r=0}^{\infty} \frac{1}{r!} = e \quad \Rightarrow \quad \sum_{r=1}^{\infty} \frac{1}{r!} = e - 1 \]

The second series, \(\sum_{r=1}^{\infty} \frac{a^r}{r!}\), is the Taylor series expansion for \(e^a\):

\[ \sum_{r=0}^{\infty} \frac{a^r}{r!} = e^a \quad \Rightarrow \quad \sum_{r=1}^{\infty} \frac{a^r}{r!} = e^a - 1 \]

### Step 5: Substitute back into the expression for \(S\)
Now substituting these results back into our expression for \(S\):

\[ S = \frac{1}{1 - a} \left( (e - 1) - (e^a - 1) \right) \]

This simplifies to:

\[ S = \frac{1}{1 - a} \left( e - e^a \right) \]

### Final Result
Thus, the final answer for the series is:

\[ S = \frac{e - e^a}{1 - a} \]