TITLE: Does $\Im(e^i+e^{e^i}+e^{e^i+e^{e^i}}\dots)$ converge?
QUESTION [15 upvotes]: 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.

REPLY [3 votes]: Update: likely the "simple inverse relation" (after eq 2) is messy. Don't know whether I can repair this
Remark: this is not an answer, but a workout which might shed more light on the whole problem
Like Oscar Lanzi remarked in his comment ("try setting the initial term...") I've looked at the behave of the iteration beginning at different starting points.
This leads to the question of the reverse function and to try, whether we can go backwards from $z_0=\exp(î)$ and see, what value $z_{-1}$ would iterate to $z_0$ and so on.
I found the following inverse function, which needs a pair of parameters at each step.
Let $$ f(z) = z + \exp(z)  \tag 1$$
then the inverse function could be written as
$$  g(z,b,k) = \log( \text{LambertW}_b(\exp(z)) + k \cdot 2 \pi î \\ k,b \in \mathbb Z \tag 2$$
[Note: the LambertW() branchindex $b$ is according to some user-implementation in Pari/GP, it might be of other sign or else way different in M'ma or Maple et al.]
Then, if we have $ z_1 = g(z_2, b,k)$ with some given $b,k$, then we have the simple inverse relation $ z_2 = f( z_1)$ .
Update: this seems only true for subsets of $(b,k)$ and $b$ and $k$ have a linear relation and that is also dependend on $z_2$. It is wrong as a general claim. The current data-example (see below) and the picture however are  correct.
Unfortunately, the opposite is not so easy. If we have $ z_2 = f(z_1)$ with a given $z_1$ then the appropriate values for $b$ and $k$ in $z_1=g(z_2,b=?,k=?)$ must empirically be determined.
It seems, in the iteration to infinity the $b$-parameter for the LambertW() governs the real part of the convergent and the $k$ branch parameter for the log() governs the imaginary part of the convergent
For instance, to find one possible precedessor for $z_0=\exp(î)$ we can choose the most simple parameter for $g()$ namely $b=0$ and $k=0$ and get
$$ z_{-1} = g(\exp(î),0,0) = -0.194208607165 + 0.469149782638 î \tag {3.1}$$
Check it:
$$    f(z_{-1}) = z_0 =  0.540302305868 + 0.841470984808 î = \exp(î) \tag {3.2} $$
Here we seem to have that $\Re (f(g(z_0,b,k)))=\Re( z_0)$ independently of the values of $(b,k)$ and only the imaginary part changes in steps of $2 \pi$.
But evaluating reversely, $ g(f(z_0),b,k) = z_0$ is only correct with unique values in $b$ and in $k$.
My guess is, that for the iterates of index, say $i=20$, when the imaginary component arrives in the convergence area and also the evolution of the real component becomes smooth, the parameters for the backward-steps $z_{19}=g(z_{20},b_{20},k_{20})$ become constant, and indeed, backwards to $z_7$ we have always $ z_{i-1} =g(z_{i},0,2)$.
Now, if we proceed from $z_7$ backwards further with the same parameters $(b,k)=(0,2)$ instead  ...
... we don't arrive at our $z_0 = \exp(î)$ but at some other value.
So let's see, what parameters we actually need when we go backwards from $z_7$ to $z_6$ and to $z_0$.
See the iterates $z_0$ to $z_9$ and the parameters $b_i$ and $k_i$ for each iteration:
                 z                      b  k                 
 --------------------------------------+--+--+----------------------    
 z_0:  0.540302305868+0.841470984808*I  0  0    --->  z_1
 z_1:    1.68413794966+2.12135398618*I  1  0    --->  z_2
 z_2:   -1.13455960996+6.71301817364*I  0  1
 z_3:  -0.842246816784+6.84701983939*I  0  1
 z_4:  -0.478178799834+7.07722157180*I  0  1
 z_5: -0.0436380449935+7.51933514381*I  0  1
 z_6:   0.270773188633+8.42353070970*I  1  1    --->  z_7
 z_7:  -0.436174626083+9.52756299919*I  0  2    --->  z_8 from here parameters 
 z_8:   -1.07926736567+9.46122892125*I  0  2          are always (0,2)

Beginning at $z_7$ we would do $z_6=g(z_7,1,1)$, $z_5=g(z_6,0,1)$, $z_4=g(z_5,0,1)$, ...

The more interesting aspect is here that we can again iterate infinitely into the backwards direction (using $g(,0,2)$) with that constant pair of parameters $(0,2)$ and iterate seemingly towards $- \infty + 4 \pi î$
Here is the plot of the partial orbits of $30$ steps on $f()$ (color:blue) namely $z_0$ to $z_{30}$
and then the partial orbits of $63$ steps on $g( ,0,2)$ (color: gold) namely $z_{30}$ to $z_{-32}$.