Coefficient of `x^n` in `e^(e^x)` is `1/(n!)` k then k =

Options

A. `1+1/(1!) +1/(2!) +.......oo`
B. `1+(2^n)/(2!) +(3n)/(3!) +.......oo`
C. `1/(2!) -1/(3!) +1/(4!) -1/(5!)+.......oo`
D. `1+(2^n)/(2!) +(3^n)/(3!) + ....oo`

Answer

To find the coefficient of \( x^n \) in the expression \( e^{e^x} \), we will follow these steps:

### Step 1: Write the series expansion for \( e^x \)

The exponential function \( e^x \) can be expressed as a power series:
\[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

### Step 2: Substitute \( e^x \) into \( e^{e^x} \)

Now, we substitute \( e^x \) into the expression for \( e^{e^x} \):
\[ e^{e^x} = \sum_{m=0}^{\infty} \frac{(e^x)^m}{m!} = \sum_{m=0}^{\infty} \frac{(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots)^m}{m!} \]

### Step 3: Identify the coefficient of \( x^n \)

To find the coefficient of \( x^n \) in \( e^{e^x} \), we need to consider how \( x^n \) can be formed from the series expansion. The terms contributing to \( x^n \) come from the expansion of \( (e^x)^m \).

The coefficient of \( x^n \) in \( e^{e^x} \) can be expressed as:
\[ \text{Coefficient of } x^n = \frac{1}{n!} \sum_{m=1}^{\infty} \frac{m^n}{m!} \]

### Step 4: Simplify the expression

The sum \( \sum_{m=1}^{\infty} \frac{m^n}{m!} \) can be simplified. This sum is known to be equal to \( e \) times the \( n \)-th Bell number \( B_n \). However, for our purposes, we can express it as:
\[ \sum_{m=1}^{\infty} \frac{m^n}{m!} = 1 + \frac{2^n}{2!} + \frac{3^n}{3!} + \ldots \]

### Step 5: Final expression for \( k \)

Thus, we can express the coefficient of \( x^n \) in \( e^{e^x} \) as:
\[ \text{Coefficient of } x^n = \frac{1}{n!} \left( 1 + \frac{2^n}{2!} + \frac{3^n}{3!} + \ldots \right) \]

Comparing this with the given expression \( \frac{1}{n!} k \), we find:
\[ k = 1 + \frac{2^n}{2!} + \frac{3^n}{3!} + \ldots \]

### Conclusion

Thus, the value of \( k \) is:
\[ k = \sum_{m=1}^{\infty} \frac{m^n}{m!} \]