express ln(x) with a = 3 as taylor series I couldn't figure out how to represent $$ln(x)$$ with a=3
as Taylor series in summation form.  
 A: Notice that
$$\begin{align}\ln(1+x)&=\int_0^x\frac1{1+t}dt\\&=\int_0^x\sum_{n=0}^\infty(-t)^ndt&\text{geometric series}\\&=\sum_{n=0}^\infty(-1)^n\int_0^xt^ndt\\\color{purple}{\ln(1+x)}&\color{purple}{=\sum_{n=0}^\infty\frac{(-1)^n}{n+1}x^{n+1}}\end{align}$$

$$\begin{align}\ln(k+x)&=\ln\left(k\left(1+\frac xk\right)\right)\\&=\ln(k)+\color{purple}{\ln\left(1+\frac xk\right)}\\&=\ln(k)+\color{purple}{\sum_{n=0}^\infty\frac{(-1)^n}{n+1}\left(\frac xk\right)^{n+1}}\\\ln(k+x)&=\ln(k)-\sum_{n=1}^\infty\frac{(-1)^n}{nk^n}x^n\\\end{align}$$
Let $k+x\to x$ and see that we get the Taylor expansion of $\ln(x)$ centered around $k$:
$$\ln(x)=\ln(k)-\sum_{n=1}^\infty\frac{(-1)^n}{nk^n}(x-k)^n$$
A: $$ln(3)+\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{3^{n}n}(x-3)^{n}$$
A: Let $g(x)=\ln(x+3)$
and $ \;f(x)=\ln(x)=g(x-3)$.
$$g(x)=\ln(3)+\ln(1+\frac{x}{3})$$
$$=\ln(3)+\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}\frac{x^k}{3^k}+O(x^n)$$
A: We have that
$${\displaystyle \log(1+x)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}} \quad {\text{ for }}|x|<1}$$
let $x\rightarrow \frac{x}{3}$
$${\displaystyle \log(1+\frac{x}{3})=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n3^n}} }$$
$${\displaystyle \log(\frac{x+3}{3})=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n3^n}} }$$
$${\displaystyle \log(x+3)=\log 3+\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n3^n}} }$$
let $x\rightarrow (x-3)$
$${\displaystyle \log(x)=\log 3+\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {(x-3)^{n}}{n3^n}} }$$
