What is the infinite sum of $a^{b^x}$?

Question

What would $$\sum^{\infty}_{n=0}(1/2)^{4^n}$$ be and how to determine it?

Note that is not a typo, it is of the form $a^{b^x}$ if it were $(1/2)^{4n}$ it would of course be trivial and could be treated using the geometric series summation formula $1/(1-r)$ with $r$ being $1/16$.

I can see this converges by the ratio test. My issue is working out its sum, more for fun really.

It expands to $(1/2) + (1/2)^{4} + (1/2)^{4^2} + \ldots + (1/2)^{4^n}$, and there doesn't seem to be anything simple to do. I have attempted to look for analogies by treating it as a function and integrating, but it doesn't seem expressible with elementary functions.

Is this a problem that cannot be tackled by elementary methods, (the only methods I currently have at my disposal)? What things should I study to be able to handle these kind of sums?

Please give some context on where this question comes from and what do you already know/your thoughts on the problem — Belgi, Aug 29 '14 at 20:20
And are you sure it's about $(\frac12)^{4^n}$ or rather $((\frac12)^4)^n$? — Frunobulax, Aug 29 '14 at 20:20
Do you actually have to determine it or check if it converges? The latter is trivial as $\sum_i (1/2)^i$ converges. — Batman, Aug 29 '14 at 20:21
Both wolfram alpha and sagemath is unable to find a closed form solution, so I doubt one exists. — Alice Ryhl, Aug 29 '14 at 20:25
Thanks for your comments. Hopefully my edits have clarified my question — user2321, Aug 29 '14 at 20:47
This is an example of a lacunary series; I learned a bit about them upon asking a question at here at MSE and another at MathOverflow. Both of those question threads have some useful discussion and links to literature. One thing I can immediately say based on what I learned there: since $1/2$ is algebraic, the series you propose is transcendental. — Semiclassical, Aug 29 '14 at 23:19

score 1 · Accepted Answer · answered Aug 30 '14 at 02:02

By asking what the infinite sum is, would you like to write it down as a decimal (using base 10)?

Review of decimals:

A decimal $.a_1a_2a_3...$ really is itself an infinite series: $$\sum^{\infty}_{n=1}\frac{a_n}{10^{n}}$$

We are comfortable with such series at least partly because we know exactly how to interpret the accuracy of such an expression: If $a$ is the above decimal and we approximate it by using the first $n$ digits, then we are within $(.1)^{n}$ of the real value of $a$.

"Evaluating" the given series:

Even if there isn't a nice way to express your given series, all is not lost; we need only know how accurate given approximations are. In fact, we can compare the tail of the given series with a geometric series: for all $k>0$, $$\sum^{\infty}_{n=k}(1/2)^{4^n} < \sum^{\infty}_{n=4^{k}}(1/2)^{n} = 2 \times (1/2)^{4^{k}}.$$

In other words, we have an upper bound on the error produced by the partial sum $\sum^{k-1}_{n=0}(1/2)^{4^n}$. And this upper bound approaches zero very very quickly: for $k = 5,$ we have the following inequality: $2 \times (1/2)^{4^{k}} < (.1)^{307}$. That means that if you evaluate $\sum^{4}_{n=0}(1/2)^{4^n}$, you will be accurate to more than 300 decimal places! It begins like this: $0.562515258789062$.

Finally, note that you are already given the binary expansion of this number by the series itself: $.10010000000000010000000000000001...$ (The set of indices where the $1$'s appear is $\{4^{n} | n = 0,1,2,...\}$.)

Thanks for the great answer. – user2321 Aug 30 '14 at 20:36 — user2321, Aug 30 '14 at 20:36

What is the infinite sum of $a^{b^x}$?

1 Answers1

Review of decimals:

"Evaluating" the given series: