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In this question, I considered a double-sequence $(x_k,y_k)$ whose first term was shown to be $x_k= k\operatorname{sgn}(\cos(k))$. I was interested when the quantity $X_n=\sum_{k=1}^n x_k$ or a corresponding sum $Y_n=\sum_{k=1}^n y_k$ were equal to zero. However, in this post I am interested in the asymptotics of $X_n$, particularly if it can be suitably renormalized in the limit as $n\to\infty$.

It's interesting to compare $X_n$ to series whose signs are periodic; $\operatorname{sgn}(\cos(k))$ is not periodic since $\pi$ is irrational. Still, using Abel summation, we can renormalize several related series:

$$\text{period 1: }\lim_{z\to 1^-}\sum_{k=1}^{\infty}k(-1)^k z^k = \frac{-1}{4}$$

$$\text{period 2: }\lim_{z\to 1^-}\sum_{k=1}^{\infty}k(-1)^{k(k+1)/2} z^k = \frac{-1}{2}$$

$$\text{period 3: }\lim_{z\to 1^-}-\sum_{k=1}^{\infty}k \cos\left(\left\lfloor\frac{k-1}{3}\right\rfloor \pi\right) z^k = \frac{-3}{4}$$

$$\text{period 4: }\lim_{z\to 1^-}-\sum_{k=1}^{\infty}k \cos\left(\left\lfloor\frac{k-1}{4}\right\rfloor \pi\right) z^k = -1$$

In fact, it's not terrible to show that if $(a_{k,n})$ is a sequence with $n$ blocks of $1$ followed by $n$ blocks of $-1$, then $\lim_{z\to 1^-}-\sum_{k=1} k a_{k,n}z^k = -n/4$. Again, the sequence $\{\operatorname{sgn}(\cos(k))\}$ does not display any obvious patterns, but perhaps this analysis could show that a suitable renormalization of $X_n$ exists. Disregarding the sign operator, the related series $\sum k\cos(k)$ is also Abel-summable:

$$\lim_{z\to 1^-}\sum_{k=1}^{\infty}k \cos\left(k\right) z^k = \frac{-1}{4}\csc^2\left(\frac{1}{2}\right)$$

I mention Abel summation because I don't believe $X_n$ is Cesaro summable. This was surprising to me, because looking at the Cesaro means $(C,\alpha)_n = \sum_{j=0}^{n} \frac{\binom{n}{j}}{\binom{n+\alpha}{j}}x_j$ for $\alpha=1,2,3,4$ showed strange behavior near $n=400$. Below is a picture of $(C,1)_n$ for $1\le n\le 1000$:

It's possible this spike is due to rounding error, but I'm not certain. At any rate, I think Abel summation is the appropriate approach. All this leaves me to believe that there is a proper normalization for $X_n$, even if the behavior of $\operatorname{sgn}(\cos(k))$ is subtle.

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  • $\begingroup$ I don't know if it helps, but sgn(sin(x)) can be expanded as an infinite Fourier series (sin(x) + sin(3x)/3+...) $\endgroup$ Commented Jul 13 at 20:01
  • $\begingroup$ @mikado true; more to the point, $\operatorname{sgn}(\cos(k)) = \frac{4}{\pi}\sum_{j=0}^{\infty}\frac{\cos((2j+1)k)}{2j+1}(-1)^j$. I'm not sure how that would help, maybe by using Abel summation (the discrete of integration by parts), one could analyze the convergence? $\endgroup$ Commented Jul 13 at 20:41
  • $\begingroup$ Might you generalise it by considering k sgn(cos(k t)) as a Fourier series? Then sum each harmonic separately? If that works, take the value for t=1? $\endgroup$ Commented Jul 13 at 22:23
  • $\begingroup$ One can also express $sgn(cos(k)) = \frac{atan(e^{i k})+atan(e^{-i k})}{2}$ $\endgroup$ Commented Jul 13 at 23:39
  • $\begingroup$ Evaluating the sum $f_N=\sum\limits_{k=1}^N k \operatorname{sgn}(\cos(k))$ and $\frac{F_N}{K}$ there seems to be a sort of periodic phenomenon with a period of approximately $360$ and $\left|\frac{F_N}{K}\right|<3$ in the range $0<K\le 5000$. $\endgroup$ Commented Jul 18 at 1:46

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Noting the comment by OP, $f(k) = \operatorname{sgn}(\cos(k)) = \frac{4}{\pi}\sum\limits_{j=0}^{\infty}(-1)^j \frac{\cos((2j+1)k)}{2j+1}$, which is the Fourier Series expansion, we are interested in the regularization of $S = \sum\limits_{k=1}^{\infty} \frac{4k}{\pi}\sum\limits_{j=0}^{\infty}(-1)^j \frac{\cos((2j+1)k)}{2j+1}$.

One difficulty of analyzing the sum is that it either involves a non-smooth $\operatorname{sgn}()$ frunction in its original form or it is a double summation in the modified form.The main step I introduce in my answer is to convert this double summation into a single summation with smooth functions so that this problem can be analyzied without those distractions

I show the summary below first and later add a section for its derivation.

$$2S = \frac{-\pi^2}{4} - \frac16 + \frac{1}{\pi^2}\sum\limits_{k=1}^{\infty} \frac{\sec(k \pi^2)(1-k \pi^2 \tan(k \pi^2))}{k^2}$$.

Since we know that $\pi$ is irrational, we know that individual terms in the summation will never blow up. But the sum will oscillate widly and will not converge. To this end, we need reguralization of this.

I tried "Borel", "Abel", "Cesaro", "Dirichlet" and "Euler" methods of the single summation in Mathematica, but it is unable to compute any of them. I think none of these methods will work here based on that, but I don't have a theoretical proof. Of these, I only understand Cesaro method (for $\alpha=1$ translates to the average of the gradually extended partial sums as the number of terms approaches infinite. Mathematica does $\alpha=1$), which seems like not a good method for this (as OP notes).

Derivation of the single summation.

I will use one of my go-to tools, Fourier Transform for this. First let us convert the odd function $g(x) = x \operatorname{sgn}(\cos(x))$ into an even function $$E(x) = |x| \operatorname{sgn}(\cos(x))$$

Substituting the Fourier Series expansion from OP, $$E(x) = |x| \frac{4}{\pi}\sum\limits_{j=0}^{\infty}(-1)^j \frac{\cos((2j+1)x)}{2j+1}$$

Now let us take Fourier Transform of this term wise. Start with general FT identity, if $$ \mathcal{F}_k[h(x)] = H(k)$$, $$ \mathcal{F}_k[h(x) \cos(a x)] = \frac{H(k - \frac{a}{2\pi})+H(k + \frac{a}{2\pi})}{2}$$

Also note that $$\mathcal{F}_k(|x|) = -\frac{1}{2\pi^2 k^2}$$

So $$\mathcal{F}_k[E(x)] = \frac{4}{\pi}\frac{-1}{2\pi^2}\frac12 \sum\limits_{j=0}^{\infty} \left\{ \frac{(-1)^j}{2j+1} \left( \frac{1}{\left(k - \frac{2j+1}{2\pi}\right)^2} + \frac{1}{\left(k + \frac{2j+1}{2\pi}\right)^2}\right) \right\}$$

I let Wolfram FullSimplify to get $$\mathcal{F}_k[E(x)] = \frac{-1+\sec(k \pi^2)(1-k \pi^2 \tan(k \pi^2))}{2k^2\pi^2}$$

Now Poisson summation says that $$2S = \sum\limits_{k=-\infty}^{\infty}\mathcal{F}_k[E(x)] $$

$$2S = \lim\limits_{k \to 0} \left\{\frac{-1+\sec(k \pi^2)(1-k \pi^2 \tan(k \pi^2))}{2k^2\pi^2}\right\} + 2\sum\limits_{k=1}^{\infty}\frac{-1}{2k^2\pi^2} + 2 \sum\limits_{k=1}^{\infty}\frac{\sec(k \pi^2)(1-k \pi^2 \tan(k \pi^2))}{2k^2\pi^2}$$

Simplifying, we get $$2S = \frac{-\pi^2}{4} - \frac16 + \frac{1}{\pi^2}\sum\limits_{k=1}^{\infty} \frac{\sec(k \pi^2)(1-k \pi^2\tan(k \pi^2))}{k^2}$$

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  • $\begingroup$ So you've gone 'out of the frying pan, and into the fire?' That is, the new series seems to defy renormalization as well... $\endgroup$ Commented Jul 24 at 13:50
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    $\begingroup$ You are absolutely right. If at all you were either struggling because of the non-smooth nature of sign function or with a double summation, this is a way to demonstrate that problem without those distractions. $\endgroup$ Commented Jul 24 at 13:58
  • $\begingroup$ It's not immediately clear to me that $S$ is reflective of the original sum. I am at work and can't rigorously check things as much as I'd like, but if you expand the derivation as you did in the linked post, I would probably award this the bounty. $\endgroup$ Commented Jul 24 at 14:11
  • $\begingroup$ I am getting some different results. Maybe there is a typo with a plus/minus sign, or a parity issue? $\endgroup$ Commented Jul 24 at 17:54
  • $\begingroup$ Sorry, I checked again. I am getting the same result. I did fix a couple of typos as I mentioned in the deleted comment (deleted to bot proof it) $\endgroup$ Commented Jul 25 at 2:52

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