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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.

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    $\begingroup$ Are you sure that inside $\cos^2$ it is $3^n x$ and not $3nx$? $\endgroup$ – mathcounterexamples.net Mar 2 at 15:33
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    $\begingroup$ In my print, it said $3^n$. It would be more great if it were $3nx$, so sad :( $\endgroup$ – ToBY Mar 2 at 15:35
  • $\begingroup$ Why do you think it has a closed form? Did you try to graph it (using a computer)? $\endgroup$ – metamorphy Mar 2 at 17:19
  • $\begingroup$ Considering the Weierstrass function as well as lacunary functions, I am very skeptical that this series has a closed form. $\endgroup$ – Sangchul Lee Mar 3 at 4:35
  • $\begingroup$ @metamorphy Um, actually, I'm still questionable having closed-form. Wolfram just said the series converges, and didn't show the exact value(I don't have any mathematical graph program in PC). But, I'm somehow believing that there exists closed-form, since if there isn't, there's no reason why this problem is included in my print that teacher gave. $\endgroup$ – ToBY Mar 3 at 4:36
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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 $$\sum_{n=0}^\infty (-1)^n \frac{\cos^3(3^n x)}{3^n} = \frac34 \times (-1)^0 \frac{\cos(3^0 x)}{3^0} = \frac34 \cos(x)$$

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    $\begingroup$ Using triple formula allows the series to do the telescoping. The result made by your keen observation is quite surprising to me. Thanks, sir. $\endgroup$ – ToBY Mar 4 at 13:58
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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.

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  • $\begingroup$ I agree that what I did is formal, but I want to show that there is no closed form. $\endgroup$ – Gevorg Hmayakyan Mar 4 at 3:53
  • $\begingroup$ I haven't thought about the derivative version. Thanks for that. Is there any typical method to show that there does not exist a closed-form? $\endgroup$ – ToBY Mar 4 at 11:54
  • $\begingroup$ Actually the closed forms of known elementary functions does not have such form. May be we are looking for larger scope? $\endgroup$ – Gevorg Hmayakyan Mar 4 at 19:18
  • $\begingroup$ Just to clarify the factor $1+3^{2k+1}$ is not common for any elementary function. Of course this is not proof. $\endgroup$ – Gevorg Hmayakyan Mar 4 at 20:18
  • $\begingroup$ I see. I'll then look through how we can clarify the denominator, especially the exponents. Thanks, sir. $\endgroup$ – ToBY Mar 5 at 2:42

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