Mystery about $\sum_{n\geqslant 1}2^{-n!}$

I was playing with some series when Wolfram told me that $$\sum_{n\geqslant 1}2^{-n!}=0.765625059604644775390625\color{Red}{000000000000}752316384526264\ldots$$ and my eyes obviously stopped at the red area. Twelve decimal places! Similarly, $$\begin{align} \sum_{n\geqslant 1}4^{-n!} &= 0.31274414(\cdots)1337890625\color{Red}{000000000000000000000000}565979\ldots \\ \sum_{n\geqslant 1}8^{-n!} &= 0.140628814(\cdots)625\color{Red}{000000000000000000000000000000000000}42579598\ldots \end{align}$$ where I omitted $30$ digits and $60$ digits respectively. Is it a coincidence or is there a deeper reason? Why are these numbers so well approximated?

Solution 1:

For the first, note that: $$ \begin{align} 2^{-1!} + 2^{-2!} + 2^{-3!} + 2^{-4!} &= 0.765625059604644775390625\color{red}{000000000000}000000\ldots \\ 2^{-5!} &= 0.000000000000000000000000\color{red}{000000000000}752316\ldots \end{align} $$ So the reason you get a bunch of zeros at this point is that $2^{-5!} = 2^{-120}$ is several orders of magnitude smaller than $2^{-4!} = 2^{-24}$. In particular, $2^{-24}$ has at most $24$ non-zero digits, which means all digits after the $24$th will be zero, while $2^{-120} < 10^{-36}$, which means the first $36$ digits of $2^{-5!}$ will be $0$. So adding them up leaves a gap of $12$ zeros.

Note that, although slightly hidden, this is not the first sequence of zeros that appears for the reason mentioned above, as the $7$th digit is a $0$ for the same reason: that $2^{-3!}$ is much bigger than $2^{-4!}$. $$ \begin{align} 2^{-1!} &= 0.500000000000000000000000000000000000000000\ldots \\ 2^{-2!} &= 0.250000000000000000000000000000000000000000\ldots \\ 2^{-3!} &= 0.015625\color{red}{0}00000000000000000000000000000000000\ldots \\ 2^{-4!} &= 0.000000\color{red}{0}59604644775390625\color{red}{000000000000}000000\ldots \\ 2^{-5!} &= 0.000000000000000000000000\color{red}{000000000000}752316\ldots \\ & \vdots \\ \hline \sum_{n \geq 1} 2^{-n!} &= 0.765625\color{red}{0}59604644775390625\color{red}{000000000000}752316\ldots \end{align} $$ The gaps will get bigger and bigger as you move to the right. For instance, starting at the $121$st digit, there will be a huge gap until the $\lceil\log_{10}(2^{6!})\rceil = 216$th digit.

For completeness, as pointed out in the comments by @Did, the reason that you get such gaps at all is that $2 \mid 10$ (or more precisely: all prime divisors of $2$ divide $10$), which means that the decimal expansions of $2^{-k!}$ always terminate, i.e., always have a finite number of non-zero digits in its decimal expansion. And because $n!$ grows ridiculously fast, you then get long strings of zeros.

Solution 2:

Note that $2^{-k}=5^k\cdot 10^{-k}$, so that is a number with $\approx \log_{10}5^k=k\log_{10}5\approx 0.7k$ nonzero digits and a total of $k$ digits. In other words: $2^{-k}$ starts with $\approx 0.3k$ zeroes. Your exponents $k=n!$ grow quite fast, so that sooner or later $0.3(n+1)!$ is much bigger than $n!$, thus leading to large blocks of zeroes.