Limit of geometric mean of N radii of an ellipse Is this equation correct?
$$\lim_{N \to \infty} \prod_{n=1}^N (a^2\cos^2 (2\pi n/N)+b^2\sin^2(2\pi n/N))^{1/N}=ab$$
If so, why?

The answer has been substantially revised. In particular, the previous revision quoted a wrong value for the integral $I(a,b)$, which resulted in an overall incorrect answer. I hope that this revision is correct. :-) 
The limit is $\Big(\frac{a+b}{2} \Big)^2$ and is not $ab$ (as claimed). I break down the solution into multiple steps for ease of understanding. 

(1) A Riemann sum and integral. We first convert the product to a sum by taking logs:
$$
\frac{1}{N} \sum_{n=1}^N \ln \left( a^2 \cos^2 (2\pi n/N) +  b^2 \sin^2 (2\pi n/N)  \right).
$$
This is the Riemann sum of the function $$f: [0,1] \to \mathbb R : x \mapsto \ln \left( a^2 \cos^2 (2\pi x) +  b^2 \sin^2 (2\pi x) \right) ,$$
corresponding to the uniform partition of $[0,1]$ into $N$ parts. Since $f$ is integrable (being a continuous function over a compact interval), as $N \to \infty$, this sum tends to

$$
\int_0^1 \ln \left( a^2 \cos^2 (2\pi x) +  b^2 \sin^2 (2\pi x)  \right) \ dx = 
\frac{1}{2 \pi} \int_0^{2\pi} \ln \left( a^2 \cos^2 y +  b^2 \sin^2 y  \right) \ dy .
$$
  Call this integral $I(a,b)$.


(2) Evaluating $I(a,b)$. We use the idea of differentiating under the integral sign.
$$
\begin{align*}
\frac{\partial }{\partial a} I(a,b) 
&= \frac{1}{2 \pi} \int_0^{2\pi} \frac{\partial}{\partial a} \ln \left( a^2 \cos^2 y +  b^2 \sin^2 y  \right) \ dy 
\\ &=
\frac{1}{2 \pi} \int_0^{2\pi} \frac{2a \cos^2y}{a^2 \cos^2 y +  b^2 \sin^2 y } \ dy .
\\ &=
\frac{a}{\pi} \int_0^{2\pi} g(y) \ dy ,
\end{align*}
$$
where $g(y) = \frac{\cos^2y}{a^2 \cos^2 y +  b^2 \sin^2 y }$. Since $g(y) = g(\pi + y)$, we can write this integral as $\frac{2a}{\pi} \int_0^{\pi} g(y) \ dy$. Similarly, since $g(y) = g(\pi - y)$, we can further simplify it to $\frac{4a}{\pi} \int_0^{\pi/2} g(y) \ dy$. 

(3) Evaluating $\int_0^{\pi/2} g(y)dy$. Using Wolfram|Alpha, we can find the indefinite integral 
$$
\int g(y) \ dy = \int \frac{\cos^2 y}{ a^2 \cos^2 y + b^2 \sin^2 y} \ dx = \frac{ay - b \ \arctan \Bigl( \frac{b \tan y}{a} \Bigr)}{a(a^2 - b^2)} \color{\Grey}{+ \mathrm{const}}.
$$
By plugging in the limits $0$ and $\pi/2$, we get 
$$
\int_0^{\pi/2} g(y) \ dy = \frac{ \left. ay  - b \; \arctan \Bigl( \frac{b \tan y}{a} \Bigr) \right|_{0}^{\pi/2}}{a(a^2 - b^2)}  = \frac{(a-b) \frac{\pi}{2}}{a(a^2 - b^2)} =  \frac{\pi}{2a(a+b)}.
$$
Therefore, $\frac{4a}{\pi} \int_0^{\pi/2} g(y) \ dy = \frac{2}{a+b}$. 

(4) Back to $I(a,b)$. We have $\frac{\partial}{\partial a} I(a,b) = \frac{2}{a+b}$. Similarly, $\frac{\partial}{\partial b} I(a,b) = \frac{2}{a+b}$. Combining these two observations, we conclude
$$
I(a,b) = 2 \ln (a+b) + \mathrm{const}.
$$
When $a=b=1$, the integral is $0$, which gives the constant to be $- 2\ln 2$. Therefore 

$$
I(a,b) = 2\ln \Bigl( \frac{a+b}{2} \Bigr).
$$


(5) Final answer. To get the final answer, we only need to exponentiating this answer. That is, the limit mentioned in the question is equal to $\Bigl(\frac{a+b}{2} \Bigr)^2$.
