Does $\Im(e^i+e^{e^i}+e^{e^i+e^{e^i}}\dots)$ converge? Consider the following sum (where $\Im(z)$ denotes the imaginary part of $z$)
$$\Im(e^i+e^{e^i}+e^{e^i+e^{e^i}}\dots)$$
I.e;
$$\Im(\lim_{n\to\infty}a_n)$$
$$a_1=e^i,\ \ \  a_{n+1}=a_n+e^{a_n}\ \ \ \forall n\geq1$$
I wrote up some generic python code (Try It Online), and was surprised to see its apparent convergence to $\approx9.424$
I'm specifically concerned with the imaginary part because the real part seems to diverge logarithmically.

Does this converge? If so, are there other expressions for the constant?

Perhaps the Dirichlet test might serve useful, though I don't know how to proceed.
 A: 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.
A: 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 $3\pi$.
A: 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 $3\pi$.
