On a certain series of complex numbers

Question

Is it possible that the above infinite series is equal to

?

Answer 1

Note that $i$ has periodic powers ie: $$i^1=i, i^2=-1, i^3=-i, i^4=1, i^5=i$$ Etc. Note the period of four. So due to the fact that consecutive odd and consecutive even powers are the negatives of one another we can say that the sum can be deduced simply by taking the modulo of $n$ Thus if $n$ $mod 4=1$ the sum equals $i$ If it is $2$ then the sum is $i-1$ If it is $3$ then the sum is $-1$ and if it is zero modulo $4$ then the sum is $0$

Answer 2

Your proposed sum for the series $s(n)=\frac{i^n(1+i)}{2}$ is wrong, as can easily be demonstrated by counter-example: for the $n=1$ case your sum gives $i=\frac{i-1}{2}$, which is false.

The series is a simple geometric progression, and we can use the familiar formula for the sum of a finite geometric series to obtain:

$$\sum_{k=1}^{n}i^k=\frac{i-i^{n+1}}{1-i}.$$

Answer 3

Another way to proceed than geometric series, is by noting that $$i + i^2 +i^3 + i^4 = i-1-i+1=0.$$ So for any $ k,m \in \mathbb{N},0\leq k \leq 3,m \geq 1, n = 4m+k,$ we have $$\sum_{s=1}^{4m+k} i^s = \sum_{t=0}^{m-1}i^{t}\underbrace{(i + i^2 +i^3 + i^4)}_{=0} +\underbrace{i^{4m}}_{=1}\left(\sum_{l=0}^{k}i^l\right) = \sum_{l=0}^{k}i^l %=\left\{\begin{array}{l l} 1 &\text{if } k=0 \\ 1+i &\text{if } k=1 \\ i &\text{if } k=2 \\ 0 &\text{if } k=3\end{array} \right.$$