If $x$ is of the form $\dfrac{a\pi}{2^b}$ (where $a$ and $b$ are integers) then $\cos^n (2^n x)$ will take the value 1 infinitely often. That deals with showing that the series diverges on a dense set.Jack said:Show that \(\displaystyle \sum_{n=1}^\infty \cos^n (2^n x)\) converges for a.e. x, but diverges on a dense set of x’s .
A series converges if the sum of its terms approaches a finite value as the number of terms increases. On the other hand, a series diverges if the sum of its terms does not approach a finite value.
There are several tests that can be used to determine if a series converges or diverges, such as the ratio test, the root test, or the integral test. In this case, we will be using the integral test.
The integral test is a method for determining the convergence or divergence of a series by comparing it to an improper integral. If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.
In this series, we can use the integral test by letting u = 2^n x and du = 2^n dx. This results in the series becoming ∑(n=1)^∞ (1/2^n) ∫ cos^n(u) du. By using the fact that 0 ≤ cos^n(u) ≤ 1, we can see that the integral converges, and therefore the series also converges.
This occurs because for almost every x, the terms in the series decrease rapidly enough for the series to converge. However, there are certain values of x, known as a dense set, where the terms do not decrease rapidly enough and the series diverges. This is due to the properties of the cosine function and the exponent 2^n, which result in a dense set of values where the series diverges.