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!

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
$$a_0b_0+(a_0b_1+b_0a_1)(x-1)+\cdots$$
