Represent $ f(x) = 1/x $ as a power series around $ x = 1 $ As stated on the title, my question is: (a) represent the function $ f(x) = 1/x $ as a power series around $ x = 1 $. (b) represent the function $ f(x) = \ln (x) $ as a power series around $ x = 1 $. 
Here's what I tried:
(a) We can rewrite $ 1/x $ as $ \frac{1}{1 - (1-x)} $ and thus using the series $ \frac{1}{1-k} = \sum_{n=0}^\infty k^n, |k| < 1 $, we can write that: 
$ \frac{1}{x} = \frac{1}{1 - (1-x)} = \sum_{n=0}^\infty  (1-x)^n, |1-x| < 1 $
I have a doubt because when I type "power series of 1/x when x = 1" on WolframAlpha the result is $ \sum_{n=1}^\infty (-1)^n \cdot (-1+x)^n $.
Am I wrong?
(b) Since $ (\ln (x))' = \frac{1}{x} $, all I have to do is integrate both sides of (a)' answer:
$ \int \frac{1}{x} dx = \int \sum_{n=0}^\infty (1-x)^n dx \therefore \ln(x) = \sum_{n=0}^\infty \frac{(1-x)^{n+1}}{n+1} + C $ and by putting $ x = 1 $ we get $ C =0 $ and thus $ \ln(x) = \sum_{n=0}^\infty \frac{(1-x)^{n+1}}{n+1} $.
Are my answers correct?
Really appreciate the help.
Have a good night, mates.
 A: I don't have enough reputation to comment, so here I am.
a) The your answer and the one you get from Wolfram|Alpha are identical.
$\sum_{n=0}^\infty (1-x)^n = \sum_{n=0}^\infty (-(x-1))^n = \sum_{n=0}^\infty (-(-1+x))^n = \sum_{n=0}^\infty (-1)^n (-1+x)^n$.
b) I think you have forgotten some minus signs. 
$\int (1-x)^n dx = \frac{-(1-x)^{n+1}}{n+1} + C$. 
A: $$
\begin{align}
\frac1x
&=\frac1{1+(x-1)}\\
&=1-(x-1)+(x-1)^2-(x-1)^3+\dots\\
&=\sum_{k=0}^\infty(-1)^k(x-1)^k
\end{align}
$$
(a) You are correct; your series is the same as mine, however, usually we expand in powers of $(x-a)^n$.
(b) integrating $\frac1t$ between $t=1$ and $t=x$ gives
$$
\begin{align}
\log(x)
&=\sum_{k=0}^\infty\frac{(-1)^k}{k+1}(x-1)^{k+1}\\
&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k(x-1)^k
\end{align}
$$
A: For the function $f:(0,\infty)\to(0,\infty)$ defined as:
$$f(x)=1/x$$
and given the general Taylor's expansion formula:
$$\sum_{k=0}^{\infty}\left(D^kf(a)\right){(x-a)^k\over k!}$$
First, calculate some of the derivaties:
$$D^0f(x)=1/x$$
$$D^1f(x)=-1/x^2$$
$$D^2f(x)=1\cdot2/x^3$$
$$D^3f(x)=-1\cdot2\cdot3/x^4$$
$$\vdots$$
$$D^kf(x)=(-1)^k\ k!/x^{k+1}$$
So the Taylor's expansion of $1/x$ around $a$ is:
$$\sum_{k=0}^{\infty}\left(D^kf(a)\right){(x-a)^k\over k!}=\sum_{k=0}^{\infty}\left((-1)^k\ k!\over a^{k+1}\right){(x-a)^k\over k!}$$
$$1/x=\sum_{k=0}^{\infty}{(-1)^k(x-a)^k\over a^{k+1}}$$
so for $a=1$, your answer is correc. And for the expansion of $\ln x$ should be:
$$\sum_{k=0}^{\infty}{(-1)^k(x-a)^{k+1}\over a^{k+1}(k+1)}$$
but note that for the first term $k=0$, it's false that
$$\ln a=0 \ \forall a\neq 1 $$
while the other terms are true... so you need to shift the index $k$ by $1$:
$$\ln x=\ln a+\sum_{k=1}^{\infty}{(-1)^{k\pm 1}(x-a)^{k}\over a^{k}\ k}$$
