I've been trying to determine the fractal dimension of the infinite sine series $f\sum_{n=1}^{\infty} \frac{\sin{(n^2x)^2}}{n^3} $.
OWN ATTEMPTS
My first attempt was to attempt to show that it is NOT a fractal by showing that it differentiable - and therefore that its fractal dimension is $1$.
To do so, I attempted to take it's term-wise derivative, which is $\sum_{n=1}^{\infty} \frac{\sin{(n^2x)}}{n}$ and prove that it is uniformly convergent. I was able to find (or rather have a helpful fellow user here explain to me) that the derivative is actually divergent - so no luck there.
What I did notice though is that $f\sum_{n=1}^{\infty} \frac{\sin^2{(n^2x)}}{n^{3+p}}$, the derivative can be shown to converge uniformly by use of the Weierstrass M-test. Now I have $2$ questions I suppose:
a) Does this knowledge (that if the exponent in the denominator were just infinitesimally bigger, the fractal dimension would be $1$) allow me to make any type of statement about my infinite sum - like for example that its fractal dimension approaches $1$ or so? Probably not, but I couldn't find any definite information on this.
b) Could anyone give me some pointers as to how I could possibly compute at least some type of bound? I've attempted to evaluate it numerically using box-counting & it looks to converge to a value very close to one, but I'd like to back this up analytically.
Thanks a lot in advance!
PS: Here a picture of the sequence - just in case that's helpful!
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