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?
 A: This is not an answer, but simply a verification of the integral in Srivatsan's answer.
Using $\cos^2(y)=\frac{1+\cos(2y)}{2}$ and $\sin^2(y)=\frac{1-\cos(2y)}{2}$,
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(a^2\cos^2(y)+b^2\sin^2(y)\right)\mathrm{d}y\tag{1}
$$
becomes
$$
\log\left(\frac{a^2+b^2}{2}\right)+\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+\frac{a^2-b^2}{a^2+b^2}\cos(2y)\right)\;\mathrm{d}y\tag{2}
$$
Letting $x=\frac{a^2-b^2}{a^2+b^2}$ and substituting $y\mapsto y/2$, the integral in $(2)$ becomes
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+x\cos(y)\right)\;\mathrm{d}y\tag{3}
$$
Taking the derivative of $(3)$ with respect to $x$ yields
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+x\cos(y)\right)\;\mathrm{d}y
&=\frac{1}{2\pi}\int_0^{2\pi}\frac{\cos(y)}{1+x\cos(y)}\mathrm{d}y\\
&=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{\frac{1-z^2}{1+z^2}}{1+x\frac{1-z^2}{1+z^2}}\frac{2\mathrm{d}z}{1+z^2}\\
&=\frac{1}{\pi(1+x)}\int_{-\infty}^\infty\frac{(1-z^2)\;\mathrm{d}z}{\left(1+\frac{1-x}{1+x}z^2\right)(1+z^2)}\\
&=\frac{1}{\pi}\int_{-\infty}^\infty\left(\frac{\frac{1}{x}}{1+z^2}-\frac{\frac{1}{x(1+x)}}{1+\frac{1-x}{1+x}z^2}\right)\mathrm{d}z\\
&=\frac{1}{x}-\frac{1}{x(1+x)}\sqrt{\frac{1+x}{1-x}}\\
&=\frac{1}{x}-\frac{1}{x\sqrt{1-x^2}}\tag{4}
\end{align}
$$
where $z=\tan(y/2)$.
Integrating $(4)$ gives
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(1+x\cos(y)\right)\;\mathrm{d}y=\log\left(\frac{1+\sqrt{1-x^2}}{2}\right)\tag{5}
$$
Substituting $x=\frac{a^2-b^2}{a^2+b^2}$ into $(5)$ and $(5)$ into $(2)$ yields
$$
\frac{1}{2\pi}\int_0^{2\pi}\log\left(a^2\cos^2(y)+b^2\sin^2(y)\right)\mathrm{d}y=2\log\left(\frac{a+b}{2}\right)\tag{6}
$$
A: 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 $\Bigl(\frac{a+b}{2} \Bigr)^2$.
