Does $\Im(e^i+e^{e^i}+e^{e^i+e^{e^i}}\dots)$ converge?

Solution 1:

After a handful of iterations we have reached $$a_n = -b_n + i(3\pi + \varepsilon_n)$$ with $b_n > 0$ and $\lvert \varepsilon_n\rvert < \frac{\pi}{2}$. Then $$e^{a_n} = -e^{-b_n}\cdot e^{i\varepsilon_n} = -\frac{\cos \varepsilon_n}{e^{b_n}} - i\frac{\sin \varepsilon_n}{e^{b_n}}$$ and $$a_{n+1} = a_n + e^{a_n} = -\biggl(b_n + \frac{\cos \varepsilon_n}{e^{b_n}}\biggr) + i\biggl(3\pi + \varepsilon_n - \frac{\sin \varepsilon_n}{e^{b_n}}\biggr)\,.$$ Thus $b_{n+1} > b_n$ and $$\varepsilon_{n+1} = \varepsilon_n - \frac{\sin \varepsilon_n}{e^{b_n}}$$ has the same sign as and smaller magnitude than $\varepsilon_n$. (Here we have $\varepsilon_n > 0$, but for other starting values one might reach imaginary parts slightly smaller than an odd multiple of $\pi$.)

It follows that $\varepsilon_n$ converges, and it remains to see that the limit is $0$. Suppose the limit were $\delta \neq 0$. Then for all $n$ we have $$\lvert \varepsilon_n - \varepsilon_{n+1}\rvert = \frac{\sin \lvert\varepsilon_n\rvert}{e^{b_n}} \geqslant \frac{\sin \lvert\delta\rvert}{e^{b_n}}$$ and it follows that $$\sum_{n = N}^{\infty} e^{-b_n} < +\infty\,. \tag{$\ast$}$$ Since $$\lvert b_n - b_{n+1}\rvert = \frac{\cos \varepsilon_n}{e^{b_n}} \leqslant e^{-b_n}$$ it further follows that $b_n$ converges, in particular $b_n < B$ for all $n$ and some $B$, but this contradicts $(\ast)$. Therefore $$\lim_{n \to \infty} \varepsilon_n = 0$$ follows.

Solution 2:

Not a full proof but a strong indication that

$$\lim_{n\to\infty}\Im(a_n)=3\pi$$

If the limit converges, then

$$\lim_{n\to\infty}(\Im(a_n)-\Im(a_{n+1}))=0$$ Thus, the solution should satisfy

$$\Im(z)=\Im(z+e^{iz})$$ $$\implies\Im(z)=\Im(z)+\Im(e^{iz})$$ $$\implies\Im(e^{iz})=0$$ $$\implies\sin(z)=0$$ $$\implies z=\pi n\ \ \ \forall n\in\mathbb{Z}$$

Considering the numerical estimates approach $3\pi$ (as pointed out by Stinking Bishop, J.G., and Gottfried Helms), either the series converges to $3\pi$, or somehow very slowly oscillates between attractive fixed points of the form $\pi n$. If this is true, then it's curious that despite the initialization of $a_1=e^i$, which is much nearer to $\pi n$ for $n\in\{-1,0,1,2\}$, it prefers to intially converge towards $3\pi$.

Solution 3:

We have, basically,

$S_{n+1}=S_n+\exp(S_n)$

Render $S_n=\alpha_n+i(k\pi+\epsilon_n)$. Then

$S_{n+1}=\alpha_n+i(k\pi+\epsilon_n)+\exp(\alpha_n+i(k\pi+\epsilon_n))$

$=(\alpha_n+\exp(\alpha_n)\cos(k\pi+\epsilon_n))+i((k\pi+\epsilon_n)+\exp(\alpha_n)\sin(k\pi+\epsilon_n)))$

Whereupon

$\alpha_{n+1}=\alpha_n+\exp(\alpha_n)\cos(k\pi+\epsilon_n)$

$\epsilon_{n+1}=\epsilon_n+\exp(\alpha_n)\sin(k\pi+\epsilon_n)$

What happens next depends on the parity of $k$. If $k$ is even, then in the limit of small $|\epsilon_n|$ we render $\cos(k\pi+\epsilon_n)\to 1$ and $\sin(k\pi+\epsilon_n)\to \epsilon_n$, thus:

$\alpha_{n+1}\to\alpha_n+\exp(\alpha_n)$

$\epsilon_{n+1}\to\epsilon_n(1+\exp(\alpha_n))$

This represents an instability because the $\epsilon_n$ term is multiplied by a factor greater than $1$, and worse that factor grows because $\alpha_n$ is increasing. We run away, in more ways than one, from this possibility.

If $k$ is odd, then $\cos(k\pi+\epsilon_n)\to -1$ and $\sin(k\pi+\epsilon_n)\to -\epsilon_n$, then:

$\alpha_{n+1}\to\alpha_n-\exp(\alpha_n)$

$\epsilon_{n+1}\to\epsilon_n(1-\exp(\alpha_n))$

Now the $\epsilon$ parameter is multiplied by a positive number less than $1$, allowing a stable condition. Also the $\alpha$ parameter decreases logarithmically; solution of the difference equation for $\alpha_n$ gives $\alpha_n\sim -\ln n$. Thus the stable fixed points are specifically odd multiples of $\pi$. We would expect convergence to an odd rather than even multiple of $\pi$.

There is a minor glitch in this result. Because $\alpha$ is decreasing, the multiplier on $\epsilon$ is approaching $1$, so the convergence of $\epsilon$ to zero slows down. This may explain why the numerical results converge only slowly to the stable fixed point at $3\pi$.