Taylor series of $\ln(x+2)$

Question

I try to determine the Taylor series of $\ln(x+2)$. Since I know $\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$, I suppose I can rewrite, \begin{align} \ln(x+2) &= \ln(1-(-(x+1)))=-\sum_{n=1}^{\infty} \frac{(-1)^n (x+1)^n}{n} \\ &= - x - 1 + \frac{1}{2}(x+1)^2 - \frac{1}{3}(x+1)^3 + \cdots \end{align}

But, I don't know why, I don't get the same result than wolframAlpha.

Answer 1

Your solution gives the Taylor series around $x = -1$, and wolfram gives it around $x = 0$.

http://www.wolframalpha.com/input/?i=taylor+ln%28x%2B2%29+around+x+%3D+-1

Check this out, it is the same as you suggested.

Answer 2

Note that $\ln (x+2)=\ln 2+\ln \left(1+\cfrac x2\right)$ will readily give you the same as wolfram.

If you look carefully at the first of the series representations in the link you provided, you will see that it is the same as your expression.