Convergence Of Series: Sin^2(2^k X)

Let's dive into determining whether the series $\sum_{n=1}^{\infty} a_n$ converges, where $a_n = \prod_{k=1}^{n} \sin^2(2^k x)$ and $x \in (-\infty, +\infty)$. This is a fascinating problem that combines trigonometry and series convergence, so let's break it down.

Initial Thoughts and Challenges

When we first encounter this problem, the natural inclination might be to reach for the ratio test. However, the ratio test involves evaluating the limit of $\frac{a_{n+1}}{a_n}$ as $n$ approaches infinity. In this case:

$\frac{a_{n+1}}{a_n} = \frac{\prod_{k=1}^{n+1} \sin^2(2^k x)}{\prod_{k=1}^{n} \sin^2(2^k x)} = \sin^2(2^{n+1} x)$

The problem here is that $\sin^2(2^{n+1} x)$ oscillates between 0 and 1 as $n$ goes to infinity, unless $x$ takes on specific values (like integer multiples of $\pi$). Therefore, the limit $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}$ doesn't generally exist, and the ratio test becomes inconclusive. So, we need to explore other avenues.

Exploring Special Cases

Before getting too deep into general approaches, let's consider some special cases that might give us insights:

1. If $x = k\pi$ where $k$ is an integer:

   In this scenario, $\sin(2^k x) = \sin(2^k k\pi) = 0$ for all $k \geq 0$. Consequently, $a_n = 0$ for all $n$, and the series converges trivially to 0.

2. If $x = \frac{k\pi}{2^m}$ for some integers $k$ and $m$:

   This case is interesting because at some point, one of the terms in the product will become zero. Specifically, when $k = m$, we have $\sin(2^m x) = \sin(k\pi) = 0$. Thus, for all $n \geq m$, $a_n = 0$, and the series again converges trivially.

These special cases show us that for certain values of $x$, the series converges quite easily. But what happens for other values of $x$?

A More General Approach

For a more general approach, we'll need to analyze the behavior of the terms $a_n$ more closely. Notice that $a_n$ is a product of squared sine functions. Since $0 \leq \sin^2(y) \leq 1$ for any $y$, we know that $0 \leq a_n \leq 1$ for all $n$. However, this alone isn't enough to determine convergence. We need to determine whether $a_n$ tends to zero quickly enough as $n$ increases.

Let's try to rewrite $a_n$ in a more manageable form. We can use the identity $\sin(2y) = 2\sin(y)\cos(y)$ to manipulate the product. Notice that:

$\prod_{k=1}^{n} \sin(2^k x) = \sin(2x) \sin(4x) \sin(8x) \dots \sin(2^n x)$

Multiply and divide by $\sin(x)$:

$\frac{\sin(x) \prod_{k=1}^{n} \sin(2^k x)}{\sin(x)} = \frac{\sin(x) \sin(2x) \sin(4x) \dots \sin(2^n x)}{\sin(x)}$

Now, repeatedly apply the identity $\sin(y)\cos(y) = \frac{1}{2}\sin(2y)$. We need to introduce cosine terms into the product. To do this, multiply and divide by $\cos(x)$, $\cos(2x)$, ..., $\cos(2^{n-1}x)$:

$\frac{\sin(x) \prod_{k=1}^{n} \sin(2^k x)}{\sin(x)} = \frac{\sin(x) \cos(x) \cos(2x) \dots \cos(2^{n-1} x) \sin(2x) \sin(4x) \dots \sin(2^n x)}{\sin(x) \cos(x) \cos(2x) \dots \cos(2^{n-1} x)}$

Applying the identity $\sin(y)\cos(y) = \frac{1}{2}\sin(2y)$ repeatedly, we get:

$\frac{\frac{1}{2}\sin(2x) \cos(2x) \dots \cos(2^{n-1} x) \sin(4x) \dots \sin(2^n x)}{\sin(x) \cos(x) \cos(2x) \dots \cos(2^{n-1} x)} = \frac{\frac{1}{2^2}\sin(4x) \dots \cos(2^{n-1} x) \sin(8x) \dots \sin(2^n x)}{\sin(x) \cos(x) \cos(2x) \dots \cos(2^{n-1} x)} = \dots = \frac{\frac{1}{2^n} \sin(2^{n+1} x)}{\sin(x)}$

Thus, we have:

$\prod_{k=1}^{n} \sin(2^k x) = \frac{\sin(2^{n+1} x)}{2^n \sin(x)}$

Therefore,

$a_n = \prod_{k=1}^{n} \sin^2(2^k x) = \left( \frac{\sin(2^{n+1} x)}{2^n \sin(x)} \right)^2 = \frac{\sin^2(2^{n+1} x)}{4^n \sin^2(x)}$

Now, our series becomes:

$\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{\sin^2(2^{n+1} x)}{4^n \sin^2(x)}$

Convergence Analysis

We can now analyze the convergence of this series. First, note that if $x = k\pi$ where $k$ is an integer, or if $x = \frac{k\pi}{2^m}$ for some integers $k$ and $m$, then $\sin(x) = 0$, and our expression is undefined. However, we've already established that the series converges to 0 in these cases.

Assuming $\sin(x) \neq 0$, we can use the comparison test. Since $0 \leq \sin^2(2^{n+1} x) \leq 1$, we have:

$0 \leq \frac{\sin^2(2^{n+1} x)}{4^n \sin^2(x)} \leq \frac{1}{4^n \sin^2(x)}$

The series $\sum_{n=1}^{\infty} \frac{1}{4^n \sin^2(x)} = \frac{1}{\sin^2(x)} \sum_{n=1}^{\infty} \frac{1}{4^n}$ is a geometric series with a common ratio of $\frac{1}{4}$, which converges because $\left| \frac{1}{4} \right| < 1$. Specifically:

$\sum_{n=1}^{\infty} \frac{1}{4^n} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$

Thus, $\sum_{n=1}^{\infty} \frac{1}{4^n \sin^2(x)} = \frac{1}{3\sin^2(x)}$.

By the comparison test, since $\sum_{n=1}^{\infty} \frac{1}{4^n \sin^2(x)}$ converges, the series $\sum_{n=1}^{\infty} \frac{\sin^2(2^{n+1} x)}{4^n \sin^2(x)}$ also converges.

Final Conclusion

In conclusion, the series $\sum_{n=1}^{\infty} a_n$, where $a_n = \prod_{k=1}^{n} \sin^2(2^k x)$, converges for all $x \in (-\infty, +\infty)$. If $x = k\pi$ or $x = \frac{k\pi}{2^m}$ for integers $k$ and $m$, the series converges trivially to 0. Otherwise, it converges by comparison to a geometric series.

So there you have it, guys! A detailed explanation of the convergence of this intriguing series. Hope this helps!