How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$
I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I want until the $x^4$ term. So what i found is the following:
$$ (1+e^x)^{-1} = 1 - e^x + e^{2x} + e^{3x} + e^{4x}+\cdots\tag1 $$
Then by expanding each $e^{nx}$ term individually:
$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+\cdots\tag2 $$
$$ e^{2x} = 1 + 2x + 2x^2 + \frac{3x^3}{2} + \frac{2x^4}{3}+\cdots\tag3 $$
$$ e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \frac{27x^4}{8}+\cdots\tag4 $$ 
$$ e^{4x} = 1 + 4x + 8x^2 + \frac{32x^3}{3} + \frac{32x^4}{3}+\cdots\tag5 $$
So substituting $(2),(3),(4),(5)$ into $(1)$ i get:
$$ (1+e^x)^{-1} = 1 +2x+5x^2+ \frac{22x^3}{3} + \frac{95x^4}{12} +\cdots$$
Which isn't the correct result. Am i not allowed to expand this series binomially? I've seen on this site that this is an asymptotic expansion. However, i don't know about these and i haven't been able to find much information on this matter to solve this. If someone could help me understand how to solve it and why my approach isn't correct, i would be VERY grateful. Thanks in advance.

$\begin{array}\\
f(x)
&=\frac{1}{1+e^x}\\
&=\frac{1}{2+(e^x-1)}\\
&=\frac12\frac{1}{1+(e^x-1)/2}\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n\frac{(e^x-1)^n}{2^n}\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n\frac{(x+x^2/2+x^3/6+...)^n}{2^n}\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n(x/2)^n(1+x/2+x^2/6+...)^n\\
&=\frac12\sum_{n=0}^{\infty}(-1)^n(x/2)^n(1+nx/2+...)\\
&=\frac12\left(1-\frac{x}{2}(1+\frac{x}{2}+...)+\frac{x^2}{4}(1+...)+...\right)\\
&=\frac12\left(1-\frac{x}{2}-\frac{x^2}{4} +\frac{x^2}{4}(1+...)+...\right)\\
&=\frac12\left(1-\frac{x}{2}+O(x^3)\right)\\
&=\frac12-\frac{x}{4}+O(x^3)\\
\end{array}
$
