I'm trying to multiply two Taylor series together to find the quartic Taylor polynomial about 0 of a function

$\displaystyle

(1+\frac{1}{2}x^2+\frac{1}{6}x^3+....)(x-\frac{1}{2}x^2+\frac{1}{3}x^3-...)

$

$\displaystyle

= 1(x-\frac{1}{2}x^2+\frac{1}{3}x^3-...)+\frac{1}{2}x^2(x-\frac{1}{2}x^2+...)+\frac{1}{6}x^3(x-...)

$

$\displaystyle

= x+(-\frac{1}{2})x^2+(\frac{1}{3}+\frac{1}{2})x^3+(-\frac{1}{4}+\frac{1}{6})x^4

$

$\displaystyle

= x-\frac{1}{2}x^2+\frac{5}{6}x^3-\frac{1}{12}x^4

$

The problem is when I enter the function into my computer maths program(mathcad) I get.

$\displaystyle

= x-\frac{1}{2}x^2+\frac{5}{6}x^3-\frac{1}{3}x^4

$

where am I going wrong

Oh the function by the way is

$\displaystyle

f(x)= (e^x-x)(\ln(1+x))

$