21
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ Try writing the first in binary the second in base four and the third in base 8. $\endgroup$ – Baby Dragon May 5 '13 at 17:36
  • $\begingroup$ I see, it is quite similar to $\sum_{k\geqslant 1}10^{-k!}$. However, I don't see a connection between this and the 12 zeroes in base ten in the examples above. $\endgroup$ – Ian Mateus May 5 '13 at 17:40
  • 3
    $\begingroup$ Intuitively what I think is happening is that in base two you have these large oceans of zeros. It seems to me that when one changes to base ten, a sizable proportion of the zeros will survive. The comment also tells us why these numbers have such good rational aproximations. $\endgroup$ – Baby Dragon May 5 '13 at 17:45
  • $\begingroup$ To mention a fact explained in the comments below, $\sum\limits_{n\geqslant1}k^{-n!}$ has "longer and longer bunches of consecutive zeroes" for every $k\geqslant2$, $k=2^i5^j$. $\endgroup$ – Did May 5 '13 at 20:55
  • 1
    $\begingroup$ "Cleaner and more helpful" to explain the gaps in the decimal expansion? I doubt that... (For the record, @Hagen's answer mentions the crucial arguments.) Anyway, now the explanation is in the comments, and even in the answer, so everything is fine. $\endgroup$ – Did May 6 '13 at 18:37
20
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ So the reason you get a bunch of zeros is that... is several orders of magnitudes smaller than... Sorry but I do not think this is the reason. Otherwise the same argument would apply to $\sum\limits_{n\geqslant1}7^{-n!}$, which has no two consecutive digits $0$ amongst its $110+$ first digits. $\endgroup$ – Did May 5 '13 at 20:24
  • 1
    $\begingroup$ @Did: Sure, if $2$ was coprime to $10$, we would have a problem. So yes, the reason that you may possibly get strings of zeros at all is that the decimal expansions terminate, but the reason that this gap appears at this point, and has length $12$, is what I described above. $\endgroup$ – TMM May 5 '13 at 20:36
  • 1
    $\begingroup$ The same applies to $6$ instead of $7$, and $6$ is not coprime to $10$ hence coprimality to $10$ is moot here... // If the fact that the expansion of $\frac12$ terminates is crucial (and it is), THIS fact should be mentioned as a reason in any answer (and then, the observation that the bunch-of-zeroes phenomenon occurs for $\sum\limits_{n\geqslant1}k^{-n!}$ if and only if $k\geqslant2$ with $k=2^i5^j$, would be right around the corner...). $\endgroup$ – Did May 5 '13 at 20:50
11
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.