How to show $y_k = \sum\limits_{i = 0}^{k-1} e^{-y_i}$ diverges as $k \to \infty$

Question

Consider the following series, $$y_k = \sum\limits_{i = 0}^{k-1} e^{-y_i}$$

where each $y_k$ is calculated using the difference equation,

$y_{k+1} = y_k + e^{-y_k}$, with $y_0 = 0$

I wish to show that this summation diverges as $k \to \infty$.

However, this seems to be very difficult to show.

Consider the first few terms of $y_k$:

$y_0 = 0,$

$y_1 = 1,$

$y_2 = 1+e^{-1}$

$y_3 = 1+e^{-1} + e^{-( 1+e^{-1})} $

...

Each iteration the sequence adds a small positive number. The question is: will these small positive numbers decrease to $0$, so that $y_k$ reaches a finite limit? If not, how can I show $y_k \to \infty$ as $k \to \infty$? Does anyone have a way to show this?

Add:

Simulation shows that,

$y_{100}$ =

4.6406

$y_{1000}$ =

6.9125 

$y_{10^6}$ =

13.8155

So it is increasing, but very slowly!

Answer 1

Set $x_n = e^{y_n}$ and then the recurrence becomes $x_{n+1} = x_ne^{\frac{1}{x_n}} = x_n(1+\frac{1}{x_n}+S_n)$, where $S_n = O(x_n^{-2}).$ This means, that $x_{n+1} = x_n+1+R_n,$ where $0\leq R_n\leq\dfrac{C}{x_n}$ for some independent, positive constant $C.$ Sum this up and you will obtain $x_{n}\geq n,$ which gives $y_n\geq \ln n$, hence proving the divergence of your sequence.

Note that, $y_{1000} = 6.9125...$ while the lower bound gives $\ln 1000 = 6.9077$, so it appears to be pretty tight.