Indefinite sum in $\mathbb{C}$ 
$$\sum\limits_{n=2}^{\infty} \left(\frac{\mathrm{i}}{2}\right)^n$$

Obviously it is a geometric series with an index shift. So I got:
$$\sum\limits_{n=2}^{\infty} \left(\frac{\mathrm{i}}{2}\right)^n = \frac{1}{1-\frac{\mathrm{i}}{2}}-\left(\frac{\mathrm{i}}{2}\right)^2=\frac{2}{2-\mathrm{i}}+\frac{1}{4}=\frac{10-\mathrm{i}}{8-4i}$$
I've checked this solution with WA, but it's wrong (Correct solution: $-\frac{1}{5}-\frac{\mathrm{i}}{10}$).
Any hints?
 A: If break down summations into two part real and imaginary:
$S=\sum_{n=2}^{\infty}(\frac{i}{2})^n$, break this into two infinite sums. 
1) case $n=2k$: $S_1=\sum_{k=1}^{\infty}(\frac{-1}{2^2})^k=\frac{-1}{5}$
2) case $n=2k+1$: $S_2=\frac{i}{2}\sum_{k=1}^{\infty}(\frac{-1}{2^2})^k=\frac{-1}{5}\frac{i}{2}$
then $S=S_1+S_2$.
Thanks to @lisyarus, for the explanation. 
However, the answer is still true but more complicate. 
Without complicating computations we can use general formula for power series $$\sum_{n=k}^{\infty}x^n=\frac{x^k}{1-x}\Leftrightarrow ||x||<1$$
A: Everything you did is right except subtraction. U should subtract i/2 and 1  instead of (i/2) square
A: Separate even and odd powers,
$$I=\sum_{n=2}^{\infty} (\frac{i}{2})^n = \sum_{k=1}^{\infty} (\frac{i}{2})^{2k} +\sum_{k=1}^{\infty} (\frac{i}{2})^{2k+1} =(1+\frac{i}{2}) \sum_{k=1}^{\infty} (\frac{i}{2})^{2k} = (1+\frac{i}{2}) \sum_{k=1}^{\infty} (\frac{-1}{4})^{k} $$
Now use sum of geometric series
$$\sum_{k=1}^{\infty} (\frac{-1}{4})^{k} = \frac{1}{1-(-\frac{1}{4})}-1=-\frac{1}{5}$$
Finally
$$I=(1+\frac{i}{2})(-\frac{1}{5})=-\frac{1}{5} -\frac{i}{10}$$ 
