I came up with the following approximation


I don't know too much about proving an inequality like this algebraically. I was hoping for an extremely rigorous proof of this (I would definitely appreciate names of theorems). I am just starting to self study computational number theory.

I didn't know how to prove this whatsoever. I would think of using a large finite number of iterations on a Taylor series, but I really had no clue how to use that. Thanks for any help.

An similar question type to this is the following: Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$. My wording is a bit odd in this question, so please note that both questions are very similar. (Solving mine algebraically is really the basis, though)

  • $\begingroup$ Oh yeah, could you please prove assuming you were not given a calculator. $\endgroup$
    – user285523
    Nov 24, 2015 at 1:52
  • $\begingroup$ What do you mean by the symbol $\lesssim$? $\endgroup$
    – ajd
    Nov 24, 2015 at 1:53
  • $\begingroup$ Approximately but still greater than is what that symbol means. You also have it backwards :) @AlexanderDunlap $\endgroup$
    – user285523
    Nov 24, 2015 at 1:55
  • $\begingroup$ Oh, boy... $\endgroup$
    – Lucian
    Nov 24, 2015 at 2:38

2 Answers 2


This isn't really by hand, but here is an explanation of what OP found. We can calculate the continued fraction for $\sqrt[4]{\pi}$ as $$[1;3,55,3,1,1,2,3,37,\ldots]$$ We get a good approximation in a continued fraction by stopping just before a "big" number. Hence we get a quite good estimate as $$\sqrt[4]{\pi}\approx [1;3]=\frac{4}{3}$$ We can refine this by taking more terms: $$\sqrt[4]{\pi}\approx [1;3,55]=\frac{221}{166}\approx 1.3313253\ldots$$ This explains why $\sqrt[4]{\pi}+0.002$ is close to $\frac{4}{3}$. To prove the requested bound, we need to go a bit further. The continued fractions alternate as an overestimate, followed by an underestimate. $[1;3]$ is an overestimate (for $\sqrt[4]{\pi}$), $[1;3,55]$ is an underestimate, and so on. Just a few more terms gives us: $$\sqrt[4]{\pi}>[1;3,55,3,1,1]=\frac{1555}{1168}\approx 1.3313356>\frac{4}{3}-0.002$$ Hence $\sqrt[4]{\pi}+0.002>\frac{4}{3}$.

  • 2
    $\begingroup$ It looks as if you didn't read the comment below the OPs question (-1) $\endgroup$
    – user266519
    Nov 24, 2015 at 2:13
  • 1
    $\begingroup$ Everyone's right. See, I designed this question with the wanting to solve this algebraically. As I didn't specify this very well, I will give a +1 $\endgroup$
    – user285523
    Nov 24, 2015 at 2:21
  • $\begingroup$ I have edited the question earlier $\endgroup$
    – user285523
    Nov 24, 2015 at 2:24
  • $\begingroup$ (a) The OP did not include the restriction to hand-calculation in the question itself. (I don't fault the OP, you understand, it's just an observation.) (b) This answer states upfront that it is not by hand; it is not misleading. (c) It is posted in advance of any answer that is by hand. And (d) it contributes to an understanding of the approximation. I think it is therefore worthy of a compensatory upvote. +1 $\endgroup$
    – Brian Tung
    Nov 24, 2015 at 2:42

Maybe it is worth mentioning that this follows from the well-known approximation/inequality $\frac{22}{7} > \pi$ because $(4/3)^4 > \frac{22}{7}$.

  • $\begingroup$ This proves that $\sqrt[4]{\pi}<\frac{4}{3}$, which is not what is asked here. $\endgroup$
    – vadim123
    Nov 24, 2015 at 15:21

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