The sequence $a_n=(1/2)^{(1/3)^{...^{(1/n)}}}$ doesn't converge, but instead has two limits, for $a_{2n}$ and one for $a_{2n+1}$ (calculated by computer - they fluctuate by about 0.3 at around 0.67). Why is this?

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    As to "why", intuitively (outside the answer below), consider that the $n$th term has exponent $1\over n$, but in the next term, the $1\over n$ has become $1\over n^{1 \over n+1}$, which is a jump from near $0$ to near $1$. At the following iteration, the $1 \over n+1$ portion has been sent to "near $1$" from "near $0$" by its exponent of $1 \over n+2$, leaving $1 \over n$ with a near $1$ exponent once more. This process cascades down the list in this exact even-term/odd-term fashion. – abiessu Sep 18 '13 at 20:49

1 Answers1


I reduce the problem to some results on the exponential tower $x^{x^{x^{\dots}}}$, which I haven't checked how to prove, but which I assume are probably well-known.

Notice that if we write this in the form $$(1/2)^{\dots (1/999)^x}$$ where $x = (1/1000)^{(1/1001)^{\dots (1/n)}}$, then it is enough to show that $x$ oscillates between two values. I claim that $x \le (1/1000)^{(1/1000)^{\dots (1/1000)}}$ if $n$ is odd (so that there are an even number of levels in the formula defining $x$, and the inequality sign is reversed when $n$ is even. We can show this by showing by induction on $k$ that $$(1/(n+1))\wedge(1/(n+2))\wedge \dots \wedge (1/(n+k)) \ge (1/n)\wedge (1/(n+1))\wedge \dots \wedge (1/(n+k-1))$$ when $k$ is even and a similar inequality when $k$ is odd. To do the induction we also need the fact that $$(1 / (n+1)) \wedge (1/(n+1)) \wedge \dots \wedge (1/(n+1)) \ge (1/n) \wedge (1/n) \wedge \dots \wedge (1/n)$$ when there are an even number of terms and the inequality is reversed otherwise.

Then I refer to folklore how the exponential tower $x \wedge x \wedge \dots$ does not converge when $x < e^{-e}$, and instead jumps between two values (which will be very close to $0$ and $1$ for small $x$). We know that $e^{-e} \ge 1/1000$, so the result follows.

J. J.
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