Let $$f(z) = (1-z)e^{\left(z + \frac{z^2}{2}\right)} = 1 + \sum_{n=1}^\infty a_n z^n.$$​

Which of the following is FALSE ?

Correct option is D

​$$\text{We have:} \\[10pt]f(z) = (1 - z) e^{\left( z + \frac{z^2}{2} \right)} = 1 + \sum_{n=1}^{\infty} a_n z^n. \\[10pt]\text{Differentiating } f(z): \\[10pt]f'(z) = (1 - z) \left[ (1 + z) e^{\left( z + \frac{z^2}{2} \right)} \right] + e^{\left( z + \frac{z^2}{2} \right)} (-1) \\[10pt]f'(z) = e^{\left( z + \frac{z^2}{2} \right)} \left[ (1 - z)(1 + z) - 1 \right] = -z^2 e^{z} e^{\frac{z^2}{2}}. \\[10pt]\text{Now, expand } f(z): \\[10pt]f(z) = (1 - z) e^z e^{\frac{z^2}{2}}. \\[10pt]\text{Expanding } e^z \text{ and } e^{\frac{z^2}{2}} \text{ into series:} \\[10pt]f(z) = (1 - z) \left[ 1 + \frac{z}{1!} + \frac{z^2}{2!} + \dots \right] \left[ 1 + \frac{z^2}{2} + \frac{z^4}{2! \cdot 2^2} + \dots \right]. \\[10pt]\text{From this:} \\[10pt]a_1 = -1 + \frac{1}{1!} = 0, \quad a_2 = \frac{1}{2} + \frac{1}{2} - 1 = 0. \\[10pt]\text{Thus:} \\[10pt]a_1 = a_2. \\[10pt]\text{For } n \in \mathbb{N}, \, a_n \in (-\infty, 0] \text{ since the coefficients of } z^n \text{ have positive terms } \leq \text{ negative terms.} \\[10pt]\text{The statement } \sum_{n=3}^{\infty} |a_n| < 1 \text{ is } \textbf{FALSE} \text{ because the series does not converge to less than } 1. \\[10pt]\textbf{Correct Answer: (D).}$$​​