Consider $f(z) = \dfrac{z+i}{z^{2}}$ , about $z = i$ and $-i \in D$, where $D$ - area of factorizing.
My proof :
consider $t = z-i$, so we have $h(t) = \dfrac{1}{t+i} = \dfrac{1}{t(1+i/t)} = \sum_{n \ge 0} \dfrac{i^{n}}{t^{n+1}}$, so $h'(z) = \sum_{n \ge 0} \dfrac{i^{n}(-1)(n+1)}{t^{n+2}} = \dfrac{-1}{(t+i)^{2}}$ , so we have
$(t+2i) \sum_{n \ge 0} \dfrac{i^{n}(n+1)}{t^{n+1}}$ , let $n' = -(n+1)$, so we have $\displaystyle \sum_{-\infty}^{-1} (t+2i)(-n') i^{-n'-1}t^{n'} = \sum_{-\infty}^{-1}(t^{n'+1}(-i)^{n'+1} + 2(-i)^{n'}t^{n'})(-n')$
But answer in my book $\displaystyle \sum_{-\infty}^{-1}(n+2)i^{n+1}(z-i)^n$
Maybe I have calculation problems ?
Your steps are fine. There is a slight mistake (in red) here:
$$ \dfrac{1}{t(1+i/t)} = \sum_{n \ge 0} \dfrac{({\color{red}{-}}i)^{n}}{t^{n+1}} $$ Continuing gives (note the minus) $$ \displaystyle \sum_{-\infty}^{-1} (t+2i)(-n') ({\color{red}{-}}i)^{-n'-1}t^{n'} = \sum_{-\infty}^{-1}(t^{n'+1}i^{n'+1} {\color{red}{-}} 2 i^{n'}t^{n'})(-n')\\ = \sum_{-\infty}^{-1}t^{n'+1}i^{n'+1}(-n') {\color{red}{-}} 2 \sum_{-\infty}^{-1} i^{n'}t^{n'}(-n') $$
Apply an index shift to the second sum: $n' = m + 1$ which gives for the second sum
$$ 2 \sum_{-\infty}^{-2}i^{m+1}t^{m+1}(-m-1) =\\ -2(-(-1)-1) +2 \sum_{-\infty}^{-1}i^{m+1}t^{m+1}(-m-1) =\\ 2 \sum_{-\infty}^{-1}i^{m+1}t^{m+1}(-m-1) $$ where in the second line a term for $m=-1$ was added and subtracted which is found to be zero (third line). So combining the indices with $n' = m = n$ gives
$$ \sum_{-\infty}^{-1}i^{n+1}t^{n+1}(-n -2 (-n-1))\\ = \sum_{-\infty}^{-1}i^{n+1}t^{n+1}(2+n)\\ = \displaystyle \sum_{-\infty}^{-1}(n+2)i^{n+1}(z-i)^{n \color{red}{+1}} $$ so this is almost the desired (as you wrote it) result - are you sure about the last red exponent?