Evaluating $\frac{\sum_{k=1}^{1010} i^{2k-1}}{\prod_{k=1}^{1010} i^{2k}}$, where $i$ is the imaginary unit

Question

Can somebody help me evaluate the following?

$$\frac{\sum_{k=1}^{1010} i^{2k-1}}{\prod_{k=1}^{1010} i^{2k}}$$

Where $i$ is the imaginary unit: $i^2 = -1$.

Edit: Thanks for helping with formatting Henry, Blue, and Theo Bendit!

Answer 1

Since $$i^{2k-1}={(i^2)^k}\cdot-i={-1}^k\cdot-i$$ and $$\sum_{k=1}^{1010} -1^k $$ is a alternating sum, then $$\sum_{k=1}^{1010} -1^k = 1(505) + -1(505) = 1 -1 = 0$$

Also, $i^{2k} = (i^2)^k = (-1)^k$, $$\prod_{k=1}^{1010} i^{2k} = \prod_{k=1}^{1010} (-1)^k = -1$$

Hence $$\dfrac {-i \ {\sum_{k=1}^{1010} -1^k}}{\prod_{k=1}^{1010}i^{2k}} = \dfrac {0}{-1} = 0.$$

NOTE: Credit also goes to Theo Bendit, who helped out in the comments.

Answer 2

Forget the bottom

Note this:

$\sum_{k = 1}^{1010}i^{2k - 1} =-i\sum_{k = 1}^{1010}i^{2k} = -i(-1 + \sum_{k = 0}^{1010}i^{2k}) = i -i\sum_{k = 0}^{1010}(i^{2})^{k} = i -i\sum_{k = 0}^{1010}(-1)^{k} = i -i\frac{1 - (-1)^{1011}}{1 - (-1)} = i -i\frac{2}{2} = 0 $

I hope I have not made a mistake in the accounts, please check it