Are there many different power series representation for a given function? So I have to find the power series representation for $f(x) = \ln (3-x)$.
I attempted the following:

$$\ln(3-x) =  \int {- \frac{1}{3-x} dx}$$ $$ = - \int {
 \frac{1}{1-(x-2)} dx}$$ $$ = - \int {\sum_{n=0}^{\infty}{(x-2)^n} dx}$$ $$
 = \sum_{n=0}^{\infty} {\int(x-2)^ndx}$$ $$ = \bigg(-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}\bigg)+K $$

Then if we let $x=2$, then we obtain that $K=0$.
Hence the power series representation for $f(x)$ is $-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}$, where $|x-2|<1$.
However the answer from my lecturer is given as:
$$\ln(3)-\sum_{n=1}^{\infty}{\frac{x^n}{n\cdot3^n}}$$
Am I doing a mistake? Or are there many different power series representation for a given function? Any clarification would be highly appreciated.
 A: Both series are correct.  The one from the lecture is the series expansion around $x=0$, while the one derived in the posted question is the series expansion around $x=2$.  And one could choose other arbitrary points around which to expand the function.
Using a straightforward approach we see that for $f(x)=\log(3-x)$, we have for $n>0$
$$f^{(n)}(x)=(-1)^{n+1}(n-1)!(x-3)^{-n} \tag 1$$
We will use this in Approach 2 of the expansions around both $x=0$ and $x=3$ in that which follows.

EXPANSION AROUND $x=0$
Approach 1:
Using the approach outlined in the posted question, we find that 
$$\begin{align}
\log(3-x)&=-\int_2^x \frac{1}{3-t}dt\\\\
&=-\int_2^x\frac{1}{1-(t-2)}dt\\\\
&=-\sum_{n=0}^{\infty}\int_0^x (t-2)^n\\\\
&=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}
\end{align}$$
which converges for $-1\le x<3$ and diverges otherwise.

Approach 2:
From $(1)$, we can see that $f^{(n)}(2)=(-1)^{n+1}(n-1)!(-1)^{-n}=-(n-1)!$
Therefore, we can write the series representation as 
$$\log(3-x)=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}$$
which converges for $-1\le x<3$ and diverges otherwise as expected!

EXPANSION AROUND $x=3$
Approach 1:
Using the approach outlined in the posted question, we find that 
$$\begin{align}
\log(3-x)&=\log 3-\int_0^x \frac{1}{3-t}dt\\\\
&=\log 3-\frac13\int_0^x\frac{1}{1-(t/3)}dt\\\\
&=\log 3-\frac13\sum_{n=0}^{\infty}\int_0^x (t/3)^n\\\\
&=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n}
\end{align}$$
which converges for $-3\le x<3$ and diverges otherwise.

Approach 2:
From $(1)$, we can also see that $f^{(n)}(0)=(-1)^{n+1}(n-1)!(-3)^{-n}=-\frac{(n-1)!}{3^n}$.
Therefore, we can write the series representation as 
$$f(x)=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n}$$
which converges for $-3\le x<3$ and diverges otherwise as expected!
A: Hint. Your route is OK, but you should rather start with
$$
\ln(3-x) =  -\int_0^x { \frac{1}{3-t} dt}+\ln 3
$$ then follow the same path to obtain the right answer.
