Showing there's no closed-form: $\sum_{n=0}^\infty(-1)^n\frac{\cos^2({3^nx})}{3^n}$

Question

Problem_

Compute $$\sum_{n=0}^\infty(-1)^n\frac{\cos^2({3^nx})}{3^n}$$

The problem is pretty simple, but it was hard for me to segregate into the partial fractions(I wanted to make a form of telescoping).

Hmmmm... My attempts were: $$\sum_{n\ge0}(-1)^n\frac{\cos^2({3^nx})}{3^n}=\sum_{n\ge0}(-1)^n\frac{1+\cos(2\cdot3^nx)}{2\cdot3^n}={1\over2}\sum_{n\ge0}\left(-{1\over3}\right)^n+\Re \sum_{n\ge0}\frac{(-1)^ne^{i\cdot2\cdot3^nx}}{2\cdot3^n}$$

From here, could you please suggest me the idea in order to continue the calculation? I still cannot solve the series $$\sum_{n\ge0}\frac{(-1)^ne^{i\cdot2\cdot3^nx}}{2\cdot3^n}$$ because there is another exponents in the exponents of the natural constant $e$. I'm also pleasure to have a hint in a different perspective. Thanks for your interest.

[EDIT_1] I surely think that there must be some typo on the given series - for example, mistyping $\pi$ as $x$ as SangchulLee and DougM mentioned through the comments, or the location of $n$(such as $3nx\rightarrow3^nx$). But I suddenly wanted to deeply focus on this series, and I just started to doubtful about the existence of closed-form of it. Furthermore, just for the curious of math, if there's no closed-form, I want to prove that.

[EDIT_2] It's also welcome to suggest another possible typo. I'm still waiting the various opinions, suggestions, ideas, and creative solutions of the series. Besides, I'm also wondering whether there is a typical method to prove that the given series has no closed-form.

[EDIT_3] Can we evaluate the series with exponents in the denominator?

I recommend to skim what I've discussed so far. You don't have to reply all the questions. Thanks for your interest one more time.

Answer 1

Comment extended to a "not an answer" answer per request.

There is another possible form of typo $$\sum_{n=0}^\infty (-1)^n \frac{\cos^{\color{red}{3}}(3^n x)}{3^n}$$ which sums to a closed form.

Start from the triple angle formula for cosine, $$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta \quad\iff\quad\cos^3\theta = \frac34\left[\cos\theta + \frac{\cos(3\theta)}{3}\right]$$ We have $$\begin{align} (-1)^n\frac{\cos^3(3^n x)}{3^n} &= (-1)^n \frac34\left[\frac{\cos(3^n x)}{3^n} + \frac{\cos(3^{n+1}(x)}{3^{n+1}}\right]\\ &= \frac34\left[ (-1)^n \frac{\cos(3^n x)}{3^n} - (-1)^{n+1} \frac{\cos(3^{n+1} x)}{3^{n+1}} \right]\end{align} $$ This allows us to turn the sum into a telescoping sum. The end result is $$\sum_{n=0}^\infty (-1)^n \frac{\cos^3(3^n x)}{3^n} = \frac34 \times (-1)^0 \frac{\cos(3^0 x)}{3^0} = \frac34 \cos(x)$$

Answer 2

Lets say $$f(x)=\sum_{n=0}^\infty(-1)^n\frac{\cos^2({3^nx})}{3^n}$$ Then $$f'(x)=-\sum_{n=0}^\infty(-1)^n\sin({2*3^nx})$$ Now $$\sin(t)=t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+...$$ with $t=2*3^nx$ $$\sin(2*3^nx)=2*3^nx-\frac{(2*3^nx)^3}{3!}+\frac{(2*3^nx)^5}{5!}-\frac{(2*3^nx)^7}{7!}+...=2*3^nx-\frac{3^{3n}(2x)^3}{3!}+\frac{3^{5n}(2x)^5}{5!}-\frac{3^{7n}(2x)^7}{7!}+...$$ As $$\sum_{n=0}^\infty(-1)^n3^{mn}=\frac{1}{3^m+1} $$ The above relation becomes: $$f'(x)=-\sum_{k=0}^\infty\frac{(-1)^k(2x)^{2k+1}}{(1+3^{2k+1})(2k+1)!}$$ Not sure if this function has a closed form in terms of elementary functions.