Problem with Discrete Parseval's Theorem I think I must be missing something obvious, but I can't for the life of me see what it is. The discrete version of Parseval's theorem can be written like this:
$$\sum_{n=0}^{N-1} |x[n]|^2 = \frac{1}{N}\sum_{n=0}^{N-1} |X[k]|^2 $$
Now, say you've got some function in time, like $x = \sin(\omega t)$. Depending on how large your value of $N$, the LHS could be arbitrarily large.
The FT of this function is a delta function at $\omega$. Everywhere else is zero, so your RHS is given by $1/N$. So how on earth does one side equal the other?! It seems to me like you've got a summation one one side and an average value on the other...
I must be missing something obvious!

It seems that you're used to a normalization convention that's different from the one used by the Wikipedia article from which you quoted the theorem. With the convention used in the article,
$$
X_k = \sum_{n=0}^{N-1} x_n \cdot\mathrm e^{-\mathrm i 2 \pi k n / N}\;,
$$
the problem you describe doesn't occur. In your example, exactly one of the $X_k$ is non-zero, and it's proportional to $N$, so the left-hand side and the right-hand side are both proportional to $N$.
