Lower and upper bounds for $\sum_{k=1}^\infty x^{2^k}$

Question

How can I calculate the below series? I don't have any idea how to tackle this series. $$\sum_{k=1}^\infty x^{2^k}$$ I would guess most likely there isn't a closed form for this expression. In which case, are there are useful lower and upper bounds?

Edit: My question is not a duplicate of the below question since I am asking for lower/upper bounds while the linked question specifically asks for a closed form. The answers in the linked question do not provide any lower/upper bounds.

To also add a weaker question, is the limit of the above function analytic on $(0,1)$?

Ordinary generating function of powers of 2

Answer 1

Not an answer to your question, I believe that no closed form exists, not any that I know of. But here's an interesting result from complex analysis regarding this series known as Hadamard Gap theorem -
"If sequence of exponents $(p_0,p_1,p_2,\cdots)$in power series $g(z)=\sum_{k=0}^{\infty} c_{p_k}z^{p_k}$ grows atleast as fast as geometric series then it cannot be analytically continued beyond its convergence disk."
There's also a more improved result called "Fabry gap theorem" which replaces the geometric sequence by any linear sequence in the above mentioned result.
Hence this series cannot be continued analytically beyond its convergence disk.

Answer 2

While I don’t know about a closed form, I think you can take the area under the upper bounding curve $x^{2^{k-1}}$ and the lower bounding curve $x^{2^k}$ as the upper and lower bounds for approximation. More clearly: $$\int_1^{\infty}x^{2^{k}}dk\leq\text{your sum} \leq \int_1^{\infty}x^{2^{k-1}}dk$$ See this graph

WolframAlpha says that the integrals can be evaluated using the exponential integral function. As I know nothing about that, I will not speculate about the details. But, I think that you can take the average of the two bounding integrals as an approximation for your summation.

Answer 3

Long comment. Substituting $x = \exp(-2^{-s})$, $s \in \mathbb{R}$, we get

$$ \sum_{k=1}^{\infty} x^{2^k} = \sum_{k=1}^{\infty} \exp(-2^{k-s}). $$

Let $f(s)$ denote the sum above. By noting that the function $s \mapsto \exp(-2^{-s})$ behaves like a unit step function,

it is reasonable to expect $f(s)$ to behave like $\operatorname{relu}(x) := \max\{0, x\}$. Indeed, the graph below compares the graph of $f(s)$ with $\operatorname{relu}(s)$:

If we believe that $f'(s)$ is increasing (which seems true based on numerical experiments though I have no idea how to prove it now), we obtain a lower bound

$$ f(s) \geq \max\{0, s - C\}, $$

where $C$ is the constant defined by

\begin{align*} C &:= \lim_{s\to\infty} (s - f(s)) \\ &= \sum_{k=0}^{\infty} \left[ 1 - \exp(-2^{-k}) \right] - \sum_{k=1}^{\infty} \exp(-2^{k}) \\ &\approx 1.3327473824328992250\ldots \end{align*}

Plugging $s = -\log_2(-\log x)$, this lower bound becomes

$$ \sum_{k=1}^{\infty} x^{2^k} \geq \max\left\{0, -\frac{\log(-\log x)}{\log 2} - C \right\}, $$

which is actually tight as $ x \uparrow 1$.