Take any $a\in\Bbb C$. For $j\neq -a$, we note that $\begin{align}z+j &= z-a+j+a\\ &= (j+a)\left(\frac{z-a}{j+a}+1\right)\\ &= (j+a)\left(1-\left(-\frac{z-a}{j+a}\right)\right).\end{align}$ Note that it is precisely in the disk $|z-a|<|j+a|$ that we have $\left|-\frac{z-a}{j+a}\right|=\frac{|z-a|}{|j+a|}<1.$ Since $\frac{1}{1-w}=\sum_{k=0}^\infty w^k$ for $|w|<1$, then letting $w=-\frac{z-a}{j+a}$, we have in the disk $|z-a|<|j+a|$ that
$\begin{align} \frac{1}{z+j} &= \frac{1}{j+a}\frac{1}{1-w}\\ &= \frac{1}{j+a}\sum_{k=0}^\infty w^k\\ &= \frac{1}{j+a}\sum_{k=0}^\infty \frac{(-1)^k(z-a)^k}{(j+a)^k}\\ &= \sum_{k=0}^\infty\frac{(-1)^k}{(j+a)^{k+1}}(z-a)^k,\\ \end{align}$
giving us the Taylor series for $\dfrac{1}{z+j}$ about $z=a$ (where $j\neq -a$).
The above general approach works just fine for $a=1$ (which is the particular case you're considering), so long as $j\neq -1$. Of course, if $j=-1$, then we'd be trying to find the Taylor series expansion of $\dfrac{1}{z-1}$ about $z=1$--an impossible task, since $\dfrac{1}{z-1}$ fails to be defined at $z=1$.