# What is the maximum of $\sum_{k=1}^{\infty} (-1)^k(^kx)$?

During my testing of the series $$\sum\limits_{k=1}^{n} (-1)^k(^kx)$$, I found that the sum converges to two limits when $$n \to \infty$$, for $$e^{-e} \lt x \le e^{1/e}$$ and oscillates between depending on whether $$n$$ is even or odd.

Here, $$^kx$$ is tetration. The notation $$^kx$$ is the same as $$x^{x^{x^{....}}}$$, which is the application of exponentiation $$k-1$$ times. Ex. $$^3x=x^{x^x}$$.

## Questions:

$$(1)$$ What is the maximum and minimum of $$\lim\limits_{n\to \infty}\sum\limits_{k=1}^{n} (-1)^k(^kx)$$ for even $$n$$?

$$(2)$$ What is the maximum and minimum of $$\lim\limits_{n\to \infty}\sum\limits_{k=1}^{n} (-1)^k(^kx)$$ for odd $$n$$?

Edit 1:

Also during my testing in PARI, I observed that the sum seems to converge to two values only in the domain of $$e^{-e} \lt x \le e^{1/e}$$. I think the reason for this maybe is that, since $$^{\infty}x$$ converges only for $$e^{-e} \lt x \le e^{1/e}$$, the sum also converges for the same domain. I would appreciate if someone could explain why the sum converges only for $$e^{-e} \lt x \le e^{1/e}$$.

Edit 2:

With the help of user Vepir, I was able to plot $$\lim\limits_{n\to \infty}\sum\limits_{k=1}^{n} (-1)^k(^kx)$$ for both even and odd $$n$$.

## Even $$n$$: ## Odd $$n$$: ## Observations from graphs:

$$(i.)$$ $$x=e^{-e}$$ is the maximum for $$\lim\limits_{n\to \infty}\sum\limits_{k=1}^{n} (-1)^k(^kx)$$ for both even and odd $$n$$ when $$e^{-e} \lt x \le e^{1/e}$$.

$$(ii.)$$ $$x=1$$ is the minimum for $$\lim\limits_{n\to \infty}\sum\limits_{k=1}^{n} (-1)^k(^kx)$$ for even $$n$$ when $$e^{-e} \lt x \le e^{1/e}$$.

$$(iii.)$$ $$x=e^{1/e}$$ is the minimum for $$\lim\limits_{n\to \infty}\sum\limits_{k=1}^{n} (-1)^k(^kx)$$ for odd $$n$$ when $$e^{-e} \lt x \le e^{1/e}$$.

Now how can we prove any of the three claims above?

• A quite common notation for such a power tower is "$x\uparrow \uparrow k$" – Peter Jul 26 at 14:49
• Letting the limit for even $n$ be $m$ and for odd $n$ be $p$, $m-p = ^\infty x$. – automaticallyGenerated Jul 26 at 15:32

First of all, $${}^\infty x$$ converges for $$e^{-e}\color{red}{\leqslant}x\leqslant e^{1/e}$$ (a good reference here is "Exponentials Reiterated" by R.A.Knoebel, mentioned in some posts on this site). But, indeed, both $$s_{2n-1}(x)$$ and $$s_{2n}(x)$$, where $$s_n(x)=\sum\limits_{k=1}^{n}(-1)^k({}^k x)$$, converge only for $$e^{-e}. To see why, let $$e^{-e}\leqslant x\leqslant e^{1/e}$$, $$y={}^\infty x$$ and $${}^n x=y(1-r_n)$$. Then $$r_{n+1}=1-y^{-r_n}$$ and $$\lim_{n\to\infty}r_n=0\implies\lim_{n\to\infty}\frac{r_{n+1}}{r_n}=\ln y.$$ Thus, when $$e^{-1} (i.e. when $$e^{-e}), both sums converge (by the ratio test). The convergence at $$x=e^{1/e}$$ follows from the fact that $$r_n$$ is positive and decreasing in this case. Finally, for $$x=e^{-e}$$ $$r_{n+1}=1-e^{r_n}\implies r_{n+2}=1-e^{1-e^{r_n}}=r_n-r_n^3/6+o(r_n^3)$$ from which one obtains $$\lim\limits_{n\to\infty}(-1)^{n-1} r_n\sqrt{n}=\sqrt{6}$$ (see this question for an approach). Thus, due to divergence of $$\sum\limits_{n\geqslant 1}\frac{1}{\sqrt{n}}$$, both $$s_{2n-1}(e^{-e})$$ and $$s_{2n}(e^{-e})$$ diverge to $$+\infty$$. This also proves $$\color{blue}{(i.\!)}$$.
The $$\color{blue}{(ii.\!)}$$ and $$\color{blue}{(iii.\!)}$$ are easy. The (elementary) observation made in the article, \begin{align} x<1&\implies x<{}^3 x<{}^5 x<\ldots<{}^6 x<{}^4 x<{}^2 x; \\ x>1&\implies x<{}^2 x<{}^3 x<\ldots, \end{align} gives $$s_{2n}(x)>0[{}=s_{2n}(1)]$$ when $$x\neq 1$$, proving $$\color{blue}{(ii.\!)}$$, and $$s_{2n-1}(x)>-1[{}=s_{2n-1}(1)]$$ when $$x<1$$. Finally, for $$x>1$$, $${}^{n+1}x-{}^n x$$ is increasing, which is proven using induction on $$n$$ and $${}^{n+1}x-{}^n x={}^n x(x^{{}^n x-{}^{n-1}x}-1).$$ Thus, $$s_{2n-1}(x)$$ is decreasing (at least) for $$x>1$$, proving $$\color{blue}{(iii.\!)}$$.