Manage your 'e's

$\mathcal R = \lim_{k\to\infty} \int_0^1 \dfrac x{2^k} \underbrace{\prod_{n=1}^k \left( e^{x/2^n} + e^{-x/2^n} \right)}_{\mathcal L} \, dx$

$\begin{aligned}\mathcal L= &\prod_{n=1}^k \left( e^{x/2^n} + e^{-x/2^n} \right)\\=&\prod_{n=1}^k \left( \dfrac{e^{x/2^{n-1}} - e^{-x/2^{n-1}}}{e^{x/2^n} - e^{-x/2^n}} \right)\\&(\color{#0C6AC7}{\mathbf{A ~Telescopic ~Product}})\\=&\left(\dfrac{e^x-e^{-x}}{e^{x/2^k} - e^{-x/2^k}}\right)\end{aligned}$

$\large\begin{aligned} \mathcal R=&\displaystyle\int_0^1\displaystyle\lim_{k\to \infty}\dfrac{\mathcal{L}\mathrm{d}x}{\frac{1}{x/2^k}}\\=&\displaystyle\int_0^1\displaystyle\lim_{k\to \infty}\dfrac{(e^x-e^{-x})\mathrm{d}x}{\underbrace{\frac{e^{x/2^k}-1}{x/2^k}}_{\large 1}- \underbrace{\frac{e^{-x/2^k}-1}{x/2^k}}_{\large -1}}\\=&\Large \displaystyle\int_0^1\dfrac{(e^x-e^{-x})}{2}\mathrm{d}x\\=&\Large\left[\dfrac{e^x+e^{-x}}{2}\right]_{\large 0}^{\large 1}\\&\\=&\Large\dfrac{\overbrace{e+\frac 1e-2}^{\left(\sqrt e-\frac1{\sqrt e}\right)^2=\frac{(e-1)^2}{e}}}{2}\\=&\Large\dfrac{(e-1)^2}{2e}\end{aligned}$

$\huge\therefore~(-1)\times 2\times 2=\color{#D61F06}{\boxed{-4}}$

We begin with the identities

-1. $\sinh(2\phi ) = 2\sinh (\phi )\cosh (\phi)$ .

-2. $\cosh(\theta ) = \frac{1}{2}\left ( e^{\theta}+ e^{-\theta}\right )$

We have $\sinh(x)$

After repeatedly applying (1) k times on the above expression on each recurring hyperbolic-sine, we achieve:

$\sinh(x) = 2^{k}\sinh(\frac{x}{2^{k}})\prod_{n=1}^{k}\cosh(\frac{x}{2^{k}})$

Now we may take the limit as k approaches infinity.

$\sinh(x) = \lim_{k\to\infty} 2^{k}\sinh(\frac{x}{2^{k}})\prod_{n=1}^{k}\cosh(\frac{x}{2^{k}})$

In order to evaluate this limit, we must define a function and use l'Hospital's Rule on the $2^{k}\sinh(\frac{x}{2^{k}})$ factor.

Let (with 'a' being an arbitrary constant) $f(y) = 2^{y}\sinh(\frac{a}{2^{y}}) = \frac{\sinh(\frac{a}{2^{y}})}{2^{-y}}$

Using l'Hospital's Rule, we differentiate the top and bottom with respect to 'y' and easily evaluate the limit.

$\lim_{y\to\infty} f(y) =\lim_{y\to\infty} \frac{(-1)(\ln 2)(2^{-y})(a)\cosh (\frac{a}{2^{y}})}{(-1)(\ln 2)(2^{-y})} =\lim_{y\to\infty} (a)\cosh (\frac{a}{2^{y}})= a\cosh(0) = a$

So,

-3. $\sinh(x) = \lim_{k\to\infty} x\prod_{n=1}^{k}\cosh(\frac{x}{2^{k}})$ (This is essentially Viète's formula.)

Now let us look at the problem:

Using (2) the expression inside the product may be simplified to $2\cosh(\frac{x}{2^{n}})$

$I = \lim_{k\to\infty} \int_{0}^{1}\frac{x}{2^{k}}\prod_{n=1}^{k}\left (2\cosh(\frac{x}{2^{n}})\right )dx.$

We now see that '2' is multiplied k times in product so it cancels with the 2^(-k) outside the product.

$I = \lim_{k\to\infty} \int_{0}^{1}x\prod_{n=1}^{k}\left (\cosh(\frac{x}{2^{n}})\right )dx.$

Using (3) this simplifies to a very simple integral.

$I = \int_{0}^{1}\sinh(x)dx = \cosh(1) - \cosh(0)$

To evaluate this into the correct form, we must use (2) to simplify the first term.

$I = \frac{1}{2}(e+\frac{1}{e}) -1 = \frac{e^{2}+1}{2e} - 1 = \frac{e^{2}-2e+1}{2e}$

The numerator of $I$ is a perfect square.

$I = \frac{(e-1)^{2}}{2e}$ . $A=-1$ , $B=2$ , $C=2$ .

$ABC = (-1)(2)(2) = \boxed{\textbf{-4}}$