If `|x| lt 1`, Coefficient of `x^3` in `1/(e^x (1+x))` is

Options

A. `17/6`
B. `-8/3`
C. `-11/6`
D. `11/6`

Answer

To find the coefficient of \( x^3 \) in the expression \( \frac{1}{e^x(1+x)} \) for \( |x| < 1 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{1}{e^x(1+x)} = \frac{1}{e^x} \cdot \frac{1}{1+x} \]
This allows us to handle the two parts separately.

### Step 2: Expand \( e^{-x} \)
The Taylor series expansion for \( e^{-x} \) is:
\[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots \]

### Step 3: Expand \( (1+x)^{-1} \)
The series expansion for \( (1+x)^{-1} \) is:
\[ (1+x)^{-1} = 1 - x + x^2 - x^3 + x^4 - \cdots \]

### Step 4: Combine the expansions
Now we need to multiply the two expansions:
\[ e^{-x} \cdot (1+x)^{-1} = \left(1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots\right) \cdot \left(1 - x + x^2 - x^3 + \cdots\right) \]

### Step 5: Identify the coefficient of \( x^3 \)
To find the coefficient of \( x^3 \), we consider the products of terms from both expansions that result in \( x^3 \):
1. \( 1 \cdot (-x^3) \) gives \( -1 \)
2. \( (-x) \cdot (-x^2) \) gives \( +1 \)
3. \( \frac{x^2}{2} \cdot (-x) \) gives \( -\frac{1}{2} \)
4. \( (-\frac{x^3}{6}) \cdot 1 \) gives \( -\frac{1}{6} \)

Now, we can sum these contributions:
\[ -1 + 1 - \frac{1}{2} - \frac{1}{6} \]

### Step 6: Simplify the expression
To combine these terms, we can find a common denominator, which is 6:
\[ -1 = -\frac{6}{6}, \quad 1 = \frac{6}{6}, \quad -\frac{1}{2} = -\frac{3}{6}, \quad -\frac{1}{6} = -\frac{1}{6} \]
Now, summing these:
\[ -\frac{6}{6} + \frac{6}{6} - \frac{3}{6} - \frac{1}{6} = 0 - \frac{4}{6} = -\frac{2}{3} \]

### Final Result
Thus, the coefficient of \( x^3 \) in \( \frac{1}{e^x(1+x)} \) is:
\[ -\frac{2}{3} \]