Write as a power series

Question

I've got to write $\sum_{n=1000}^{\infty} i^{n}\frac{z^{2n-1}}{n!}$ as a power series. However in my mind isn't it already a power series, I don't really understand what the question is asking.

I started by putting it in the form $\sum_{n=1000}^{\infty} a_n(z-a)^{2n-1}$ where a=0 and $a_n= \frac{i^{n}}{n!}$ and stated its equal to that when $a_n$=0 for all n<1999, $a_n$=1 for all odd n>1999 and $a_n$=0 for all even n>1999, but this doesn't seem write at all so i've no idea where to go from here

Answer 1

My guess is that whoever created that problem wants you to provide this answer: $\displaystyle\sum_{n=0}^\infty a_nz^n$, with$$a_n=\begin{cases}\frac{i^{\frac{n+1}2}}{\left(\frac{n+1}2\right)!}&\text{ if $n$ odd and greater than }1998\\0&\text{ otherwise.}\end{cases}$$

Answer 2

I would get rid of the $i$s like this:

$\begin{array}\\ \sum_{n=1000}^{\infty} i^{n}\frac{z^{2n-1}}{n!} &=\sum_{n=250}^{\infty}\sum_{k=0}^3 i^{4n+k}\frac{z^{2(4n+k)-1}}{(4n+k)!}\\ &=\sum_{k=0}^3 \sum_{n=250}^{\infty}i^{4n+k}\frac{z^{8n+2k-1}}{(4n+k)!}\\ &=\sum_{k=0}^3 i^k\sum_{n=250}^{\infty}i^{4n}\frac{z^{8n+2k-1}}{(4n+k)!}\\ &=\sum_{k=0}^3 i^k\sum_{n=250}^{\infty}\frac{z^{8n+2k-1}}{(4n+k)!}\\ &=\sum_{k=0}^3 i^ks_{4n+k}(z) \qquad\text{where } s_m(z)=\sum_{n=250}^{\infty}\frac{z^{2m-1}}{m!}\\ &=s_{4n}(z)+is_{4n+1}(z)-s_{4n+2}(z)-is_{4n+3}(z)\\ \end{array} $