Find the Taylor expansion of $\ln(1+x)$ around $x=0$.
I calculated: $f'(0)=1, f''(0)=-1, f'''(0)=1$, etc.
$$T(3)=f(0)+f'(0)(x-0)+f''(0)(x-0)^2+f'''(0)(x-0)^3=0 + 1x-\frac{1}{2}x^2+\frac{1}{6}x^3$$
However, the answer says the $1/6$ should be a $1/3$. Why is this? How can something cancel out if the derivatives are only multiples of $1$? And we should have $3!=6$ in the denominator?
This is a Taylor expansion about $a$: $$T_n(x)=\sum^n_{k=0}\dfrac{f^{(k)}(a)}{\boxed{k!}}(x-a)^k$$ Can you find where you went wrong ($\dfrac{1}{6}\times 2=\dfrac{1}{3}$, $f^{\prime\prime\prime}(0)=2$)?