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Not sure if the following is easy to prove or not, I obtained the result using probabilistic arguments exclusively.

Define $x_1 = 2; x_2 = 4/3; x_3 = (8\cdot 6) / (7\cdot 5); x_4 = (16\cdot 14 \cdot 12 \cdot 10) / (15\cdot 13 \cdot 11 \cdot 9)$, and so on. Prove that

  • $x_k \rightarrow \sqrt{2}$.

  • $\sum_{k=1}^\infty (\log x_k)\cdot 2^{-k}= 1/2.$

A more challenging (unsolved problem): for which values of $k$ do we have $\log x_k < (\log 2)/2$ ? Maybe one can use the Stirling formula to solve this last problem.

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  • $\begingroup$ I did some computations. Except for $k = 1$, we always have $\log x_k < (\log 2)/2$, that is, $x_k^2 < 2$. $\endgroup$ – Vincent Granville Aug 2 '19 at 1:06
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Note that, for $k>1, x_k=\dfrac{y_k/y_{k-1}}{z_k/z_{k-1}}$ where $y_k=2^k(2^k-2)\cdot\cdot\cdot2=2^{2^{k-1}}(2^{k-1})!$

and $z_k=(2^k-1)(2^k-3)\cdot\cdot\cdot1=(2^k)!/y_k$.

Thus $x_k=\dfrac{y_k/y_{k-1}}{\dfrac{(2^k)!}{(2^{k-1})!}/\dfrac{y_{k-1}}{y_k}}= \dfrac{y_k^2}{y_{k-1}^2}\cdot\dfrac{(2^{k-1})!}{(2^k)!}$.

Substituting $y_k=2^{2^{k-1}}(2^{k-1})!$ and the corresponding expression for $y_{k-1}$ and simplifying yields $$x_k=\dfrac{2^{2^{k-1}}(2^{k-1})!^3}{(2^{k-2})!^2(2^k)!}.$$

Applying Stirling's formula ${\displaystyle n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}}$ then shows $x_k\sim\sqrt2$ after a lot of cancellation

[for example, $\sqrt{2\pi}^3$ in the numerator cancels $\sqrt{2\pi}^2\sqrt{2\pi}$ in the denominator, and

$\left(\dfrac1e\right)^{2^{k-1}3}$ in the numerator cancels $\left(\dfrac1e\right)^{2^{k-2}2}$$\left(\dfrac1e\right)^{2^{k}} $ in the denominator since $2^{k-1}3=2^{k-1}+2^k$].

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  • $\begingroup$ Nice proof of the first bullet point. Does it help you to get the second one, or the inequality OP asks for? $\endgroup$ – Gerry Myerson Aug 2 '19 at 3:58
  • $\begingroup$ If I am not mistaken, the nice formula for $x_k$ does not seem to work for $k=1$. Am I wrong ? $\endgroup$ – Claude Leibovici Aug 2 '19 at 4:08
  • $\begingroup$ @ClaudeLeibovici: I think you are correct and I have inserted the qualification $k>1$; anyways, OP seems interested in asymptotics and not $k=1$ $\endgroup$ – J. W. Tanner Aug 2 '19 at 4:25
  • $\begingroup$ @GerryMyerson: thanks. it might, but I haven’t tried $\endgroup$ – J. W. Tanner Aug 2 '19 at 4:26
  • $\begingroup$ I am sorry, I did not see $k>1$ ! Nice answer and $\to +1$. $\endgroup$ – Claude Leibovici Aug 2 '19 at 4:41

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