Title says it all, this is such a classical looking series,

$$\frac1{2^{2^0}} + \frac1{2^{2^1}} + \frac1{2^{2^2}} + \ldots.$$

So, I was just wondering, is there a closed form solution known for the partial sums? If so, can someone post? I've looked around a bit and haven't found a closed form solution yet.

  • $\begingroup$ The infinite series looks suspiciously similar to constructions of transcendental numbers with series, so I'd be surprised if there was a closed form solution for partial sums unless it involved $e$ or $\pi$ or something, but even if it did, how would it work out magically to give rational number answers for the partial sums? $\endgroup$ Jul 29, 2013 at 15:26
  • $\begingroup$ Wolfram alpha gives the answer as 0.816422. It is obvious that it converges. $\endgroup$ Jul 29, 2013 at 16:51
  • $\begingroup$ @Torsten Sure, the infinite series is bounded above by 1 and can be very well-approximated by a small handful of terms because it converges so quickly, and I might even be willing to guess that there is a closed-form expression for the infinite sum. But what about the partial sums? Basic formulas parametrized by an integer don't seem to easily capture the kind of complexity of what's going on in the partial sums for this series, hence the question. $\endgroup$ Jul 29, 2013 at 17:09
  • $\begingroup$ Might be something related to Riemann Zeta. $\endgroup$ Jul 29, 2013 at 17:35
  • 1
    $\begingroup$ Normally Roth's theorem would apply and show this sort of number transcendental, but unfortunately in this case Roth's theorem doesn't quite apply: the finite sum $s_n=\frac{p_n}{q_n}=\sum_{i=0}^n 2^{-2^{i}}$ has a denominator (in simplest terms) $q_n=2^{2^n}$ and the difference between it and the infinite sum $s_\infty=\sum_{i=0}^\infty 2^{-2^{i}}$ is bounded by $2\cdot 2^{-2^{n+1}}$ (since we can bound the tail by a geometric series with ratio $\frac12$ starting at the next term). Since $q_n^2=2^{2^{n+1}}$, these 'very good' approximations are still within the Roth bounds. $\endgroup$ Jul 30, 2013 at 18:59

1 Answer 1


I doubt there's a closed expression for it, but the partial sums arise here:


and in particular the decimal expansion of the infinite series is here:


There is a (paywall) article called "Simple continued fractions for some numbers" where the problem being considered is to find the continued fraction expansion of $\sum_{n=0}^\infty \frac{1}{U^{2^n}}$. It doesn't seem to be known how to find a closed form expression for the sum however.


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