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

Question

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.

Answer 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.

Answer 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 $3\pi$.

Answer 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 $3\pi$.

Answer 4

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 $z_6=g(z_7,1,1)$, $z_5=g(z_6,0,1)$, $z_4=g(z_5,0,1)$, ...

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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 $z_{-32}$.