A calculus problem by Ryan Broder

$f(x)=\lim_{n \rightarrow \infty} \cos \left(\dfrac{x}{2^n}\right)\cos \left(\dfrac{x}{2^{n-1}}\right) \cdots \cos \left(\dfrac{x}{2}\right)$

Multiply and divide the RHS by $\sin \left(\dfrac{x}{2^n}\right)$ and combine the terms starting from the left by using the identity $\sin {2\theta}=2\sin{\theta}\cos{\theta}$ repeatedly.

Doing this, we get $\begin{aligned} f(x) &=\lim_{n \rightarrow \infty} \dfrac{1/2^n}{\sin {\left(x/2^n\right)}} \sin {x} \\ &=\dfrac{\sin{x}}{x} \end{aligned}$

Now, we need to find $N=\sum_{n=-\infty}^{\infty} \dfrac{\sin {n}}{n}$

There is a bit of a problem, as $\dfrac{\sin x}{x}$ is not defined for $x=0$ . But if we take it to be $1$ , then $N=1+2\sum_{n=1}^{\infty} \dfrac{\sin n}{n}$ since $\dfrac{\sin (-n)}{-n}=\dfrac{\sin n}{n} \ \ \ \forall \ n \in \mathbb{I}-\{0\}$

Now, recall that $1+x+x^2+ \cdots=\dfrac{1}{1-x}$

Integrate both sides of the above equality to see that $-\log (1-x)=\sum_{k=1}^{\infty} \dfrac{x^k}{k}$

Using Euler's formula , $\begin{aligned} N &=1+2\left(\Im \sum_{n=1}^{\infty} \dfrac{e^{in}}{n}\right)\\ &= 1-2\left(\Im \left\{\log(1-e^i)\right\}\right) \end{aligned}$

After using the Euler's formula and some half-angle trigonometric identities, you will find that, $1-e^i=2\sin\left(\dfrac{1}{2}\right) \exp\left(\dfrac{i(1-\pi)}{2}\right)$ Hence, $\left(\Im \left\{\log(1-e^i)\right\}\right)=\dfrac{1-\pi}{2}$ and accordingly, $N=\pi\approx3.14$ So, $\lfloor 100N \rfloor=\boxed{314}$

$\text{Note :} \ \Im \{z\} \ \text{represents the imaginary part of} \ z.$

$\huge{cos\frac{x}{2}.cos\frac{x}{2^2} ..................cos\frac{x}{2^{n}}}$

$\huge{\frac{cos\frac{\pi}{2}cos\frac{\pi}{2^{2}}..................2sin\frac{x}{2^{n}}cos\frac{x}{2^{n}}}{2sin\frac{x}{2^n}}}$

now multiplying and dividing by 2 and applying the formula $sin2x = 2sinx cosx$ it will go on reducing and thus we get

$\huge{\frac{sinx}{2^{n}sin\frac{x}{2^{n}}}}$

$\huge{ lim_{n \to \infty} \frac{sinx}{2^{n}\frac{sin\frac{x}{2^{n}}}{\frac{x}{2^{n}}}\times \frac{x}{2^{n}}}}$

thus $\huge{= \frac{sinx}{x}}$

now we can easily see $lim_{a \to +\infty} \displaystyle \sum_{x = -a}^{a} \frac{sinx}{x} = \pi$

thus answer is 314