<s>[INST] Taylor expansion of $\log(3+x)$ Using the standard result of $\log$ find the taylor expansion of $$\log{(3+x)}$$
Now I believe $$\log{(1+x)} = \log{(1+x)} = \sum^{\infty}_{n=1} \frac{(-1)^{n+1}}{n}x^{n}$$
So to find $\log{(3+x)}$ let $y=2+x$ then we are finding $\log{(1+y)}$ which we have a standard result for:
$$\log{(1+y)} = \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}y^{n} = \sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}(2+x)^{n}$$ $$ = 2 + x - \frac{(2+x)^{2}}{2} + \frac{(2+x)^{3}}{3} - \ldots$$
but the taylor expansion of $\log{(3+x)}$ is $$\log{(3+x)} = \log{(3)}+\frac{x}{3}-\frac{x^{2}}{18}+\frac{x^{3}}{81} - \ldots$$
which is drastically different to my own, what is wrong and why does this not work?
 [/INST] Write $\displaystyle 3 + x = 3\left(1 + {x \over 3}\right)$. Then 
$$\log(3 + x) = \log 3 + \log\left(1 + {x \over 3}\right) = \log 3 + \frac{x}{3} - \frac{x^2}{2\cdot 3^2} + \frac{x^3}{3 \cdot 3^3} + \cdots$$
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