Indefinite sum in $\mathbb{C}$

Question

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

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?

Answer 1

Everything you did is right except subtraction. U should subtract i/2 and 1 instead of (i/2) square

Answer 2

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 $$\sum_{n=k}^{\infty}x^n=\frac{x^k}{1-x}\Leftrightarrow ||x||<1$$

Answer 3

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

$$I=(1+\frac{i}{2})(-\frac{1}{5})=-\frac{1}{5} -\frac{i}{10}$$