Find the power series of $\frac{3x+4}{x+1}$ around $x=1$. I'm trying to find the power series of
$$
\frac{3x+4}{x+1}
$$
around $x=1$.
My idea was to use the equation
$$
\left(\sum_{n\ge0}a_n (x-x_0)^n\right)\left(\sum_{k\ge0}b_k (x-x_0)^k\right) = \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-x_0)^n \tag{1}
$$
taken from here, to solve this problem.

My attempt
I can write the numerator of the expression as
$$
3x +4 = 3(x-1)+7 = \sum_{n\ge0}a_n (x-1)^n
$$
where
$$
a_n=\begin{cases}
7 & n=0 \\
3 & n=1 \\
0 & n>1 
\end{cases}
$$
On the other hand, given $|x-1| <2$ we get
\begin{align*}
\frac{1}{x+1} = \frac{1}{2} \cdot \frac{1}{1-\left(\frac{1-x}{2}\right)} = \frac{1}{2} \sum_{k\ge0} \left(\frac{1-x}{2}\right)^n = \sum_{k\ge0}\underbrace{\frac{-1}{(-2)^{k+1}}}_{\color{blue}{b_k}} (x-1)^k
\end{align*}
And then, using equation $(1)$ we get
\begin{align*}
\frac{3x+4}{x+1} &= \left(\sum_{n\ge0}a_n (x-1)^n\right)\left(\sum_{k\ge0}b_k (x-1)^k\right)\\
&= \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-1)^n\\
&= \sum_{n\ge0}\left(a_0 b_n + a_1 b_{n-1} \right)(x-1)^n\\
&= \sum_{n\ge0}\left(7 \frac{-1}{(-2)^{n+1}} + 3 \frac{-1}{(-2)^{n}} \right)(x-1)^n\\
&= \sum_{n\ge0}\left( \frac{-7 +3(-1)(-2)}{(-2)^{n+1}}  \right)(x-1)^n\\
&= \sum_{n\ge0} \frac{-1}{(-2)^{n+1}}  (x-1)^n\\
\end{align*}
And this seems to imply that $\frac{3x+4}{x+1}  = \frac{1}{x+1}$, at least for $|x-1| <2$, which is clearly not true.
I've gone over the steps, but I don't see where my mistake is. Could anyone tell me where my solution went wrong? Thank you!
 A: Correction to your answer
Numerator:
$$
3x+4=7+3(x-1)\tag1
$$
Denominator:
$$
\begin{align}
\frac1{x+1}
&=\frac1{2+(x-1)}\tag2\\[6pt]
&=\frac12\frac1{1+\frac{x-1}2}\tag3\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{2^{k+1}}(x-1)^k\tag4
\end{align}
$$
Multiplying the series:
$$
\begin{align}
&(7+3(x-1))\left(\sum_{k=0}^\infty\frac{(-1)^k}{2^{k+1}}(x-1)^k\right)\\
&=\color{#C00}{\sum_{k=0}^\infty7\frac{(-1)^k}{2^{k+1}}(x-1)^k}+\color{#090}{\sum_{k=0}^\infty3\frac{(-1)^k}{2^{k+1}}(x-1)^{k+1}}\tag5\\
&=\color{#C00}{\frac72+\sum_{k=1}^\infty7\frac{(-1)^k}{2^{k+1}}(x-1)^k}+\color{#090}{\sum_{k=1}^\infty3\frac{(-1)^{k-1}}{2^k}(x-1)^k}\tag6\\
&=\frac72+\sum_{k=1}^\infty7\frac{(-1)^k}{2^{k+1}}(x-1)^k-\sum_{k=1}^\infty6\frac{(-1)^k}{2^{k+1}}(x-1)^k\tag7\\
&=\frac72+\sum_{k=1}^\infty\frac{(-1)^k}{2^{k+1}}(x-1)^k\tag8
\end{align}
$$
Explanation:
$(5)$: the red sum is from the $7$, the green from the $3(x-1)$
$(6)$: pull out the constant term from the red sum
$\phantom{\text{(6):}}$ substitute $k\mapsto k-1$ in the green sum
$(7)$: make the right sum look like the left sum
$(8)$: combine the sums

How I would do it
$$
\begin{align}
\frac{3x+4}{x+1}
&=3+\frac1{x+1}\tag9\\[9pt]
&=3+\frac1{(x-1)+2}\tag{10}\\[9pt]
&=3+\frac12\frac1{1+\frac{x-1}2}\tag{11}\\
&=\frac72+\sum_{k=1}^\infty\frac{(-1)^k}{2^{k+1}}(x-1)^k\tag{12}
\end{align}
$$
Explanation:
$\phantom{1}(9)$: divide the polynomials
$(10)$: put in terms of $x-1$
$(11)$: put the fraction in the form $\frac1{1+u}$
$(12)$: use the series $\frac1{1+u}=\sum\limits_{k=0}^\infty(-1)^ku^k$
$\phantom{\text{(12):}}$ and combine the $k=0$ term with $3$
A: Hint: $f(x)=\frac{3x+4}{x+1}=3+\frac{1}{x+1}$ and from here we obtain $f^{(n)}(x)=\frac{(-1)^n n!}{(x+1)^{n+1}}, n \geqslant 1.$
Now, knowing derivative, Taylor series will be .. can you finish?
Addition:
It should be
$$\frac{3x+4}{x+1}=\frac{7}{2}-\frac{1}{4}(x-1)+ \cdots$$
and
$$\frac{1}{x+1}=\frac{1}{2}-\frac{1}{4}(x-1)+ \cdots$$
And about mistake - it is 3-d line of series multiplication. There should be
$$a_0b_0+(a_0b_1+b_0a_1)(x-1)+\cdots$$
A: You know that all power series that converge to the function $f$ agree. So you can just write for $u = x - 1$:
$\begin{align*}
   \frac{3 x + 4}{x + 1}
      &= \frac{3 u + 7}{u + 2} \\
      &= 3 + \frac{2}{1 + u / 2} \\
      &= 3 + 2 \sum_{n \ge 0} (-1)^n \left( \frac{u}{2} \right)^n \\
      &= 3 + \sum_{n \ge 0} \frac{(-1)^n}{2^{n - 1}} (x - 1)^n
\end{align*}$
A: To make life simpler, let $x=y+1$ $$A=\frac{3x+4}{x+1}=\frac{3y+7}{y+2}=3+\frac 1 {y+2}$$ Let $y=2t$
$$A=3+\frac 12 \frac 1 {1+t}$$
Make the expansion around $t=0$ and go back to $y$ and then $x$
