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On a piece of coursework I am working on, we are asked to perform some calculations involving the following: enter image description here

As you can see, it is possible for $(N-1)/2$ to be fractional, e.g. in the case $N=4$. How exactly do I go about computing this sum, as I have never seen fractions being involved before.

TerryStone
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    If not explained differently, indices are positive integers, so in this case just $k=\lfloor \frac{n-1}{2}\rfloor$. Often there is the requirement added that $N$ is odd (two lines earlier). For formulas, please use MathJax. Here is a tutorial. – Dietrich Burde Apr 18 '23 at 14:46
  • If $\theta_i$ does not depend on $N$, you can take $N=2m+1$. – Ivan Kaznacheyeu Apr 18 '23 at 14:54
  • Are you able to do this sum for $N$ odd(where the index $\frac{N-1}{2}$ is an integer)? If so, then the first comment tells you that the answer for $N=2m$ is the same as the answer to $2m-1$ for all $m \geq 2$. – Sarvesh Ravichandran Iyer Apr 18 '23 at 15:00
  • If $N$ is required to be odd, then there's of course no problem. If $N$ can be even, the most natural interpretation would be that you only sum the $k$ that satisfy $k\le (N-1)/2$. However, it would be more proper for this to be explicitly specified with the floor function. – eyeballfrog Apr 18 '23 at 15:07

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