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$$\large \pi^3 \gt 31$$

Using a calculator, $\pi^3/31 \approx 1.0002$, so I thought this may be challenging to do by hand.

It is extremely easy with the use of any calculator, so I was wondering now:

Can you prove the above inequality without the use of calculator or advanced computation in an elegant manner?

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    $\begingroup$ 3.1415^3>31 (by hand) $\endgroup$ – sinbadh Jan 12 '16 at 23:14
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    $\begingroup$ There are all sorts of questions that begin with "prove without using a calculator." In my opinion, posing the question for the sake of the puzzle is reason enough, and in any case the question is clearly of some relevance to the OP. $\endgroup$ – Elliot G Jan 12 '16 at 23:41
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    $\begingroup$ I edited. It's a fair question plagued by some red herrings, like the fact that none of us care that 31 is prime or find it especially provocative. Edited out. $\endgroup$ – djechlin Jan 13 '16 at 1:25
  • $\begingroup$ @djechlin Great edit. Thank you $\endgroup$ – user266519 Jan 13 '16 at 3:42
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The simplest way is to use the series:

$$\frac{1}{1^6}+\frac{1}{3^6}+\frac{1}{5^6}+...=\sum\limits_{k=1}^{\infty}\frac{1}{(2k-1)^6}=\frac{\pi^6}{960}$$

Now we want to prove that $$\frac{\pi^6}{960}>\frac{31^2}{960}$$ which means that we need to prove

$$960\sum\limits_{k=1}^{\infty}\frac{1}{(2k-1)^6}>31^2=961$$

However it is

$$960\sum\limits_{k=1}^{\infty}\frac{1}{(2k-1)^6}>960\left(\frac{1}{1^6}+\frac{1}{3^6}\right)=960+\frac{960}{3^6}$$

And now $\frac{960}{3^6}>1$ meaning $960>3^6$ because $320>3^5=9^2 \cdot 3=81 \cdot 3 = 243$

(Very minimal calculations involved, just to prove the point, although it was almost obvious once we got there.)

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  • $\begingroup$ Could be streamlined, all you really need is $3^6=729<960$ so$$\pi^6>960\Bigl(1+\frac1{960}\Bigr)=31^2\ .$$Very nice all the same, +1. $\endgroup$ – David Jan 13 '16 at 1:18
  • $\begingroup$ Elegant, concise, and in the spirit of the question asked. $\endgroup$ – A.S. Jan 13 '16 at 1:18
  • $\begingroup$ I tend to be detailed slightly above just marking the solution path or giving general instructions even when all details are plotted. You just do not know who is reading. Like a student, every line with its purpose. Anyway it is just 960 .vs. 961 to compare. One step off and I fail. (Not that it did not happen.) $\endgroup$ – user195934 Jan 13 '16 at 1:24
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    $\begingroup$ This is like using a jackhammer to pound in a framing nail. The facts relied on here need significant more computation than multiplication. You are just skipping over these details in the guise of known facts(but how do you know them? Because you did the computation originally). You might as well just say $\pi^3>31$ because I know that $\pi^3>31.00000001$. No computation necessary.@AlexPeter this is a complaint about the nature of the question, not about your answer, it is great, but just hides computation. $\endgroup$ – Sean English Jan 13 '16 at 3:03
  • $\begingroup$ @SE318 What is the alternative solution? How do you actually approximate $\pi$ instead? "Remembering" $3.1415\dots$ is not good enough. $\endgroup$ – A.S. Jan 13 '16 at 3:47
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We know that $\zeta(6)=\frac{\pi^6}{945}$ and $\sum_{k=1}^\infty \frac1{k^6}$ converges quite quickly. It turns out we only need $k=1,2,3$ to prove the inequality.

$$\zeta(6)=\frac{\pi^6}{945}\Leftrightarrow \pi^3=\sqrt{945\zeta(6)}\Rightarrow \pi^3>\sqrt{945\sum_{k=1}^3 \frac1{k^6}}=\frac{\sqrt{4982145}}{72}>31.$$

Calculating the last square root is not very easy but it proves the inequality.

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    $\begingroup$ That's so awkward. I'm sure! Knowing the value of $\zeta(6)$ is far easier than the first few numbers of $\pi$! $\endgroup$ – Paul K Jan 13 '16 at 1:08
  • $\begingroup$ @menag that's kind of a fair point but you can get $\zeta(2n)$ in a few pages of hand-written math proof, so perhaps if a few more digits of pi were required this would seem less of a stretch. $\endgroup$ – djechlin Jan 13 '16 at 1:22
  • $\begingroup$ @menag conversely if you can get $\zeta(2n+1)$ in any number of pages of proof for any $n > 0$ you will enjoy fame and fortune. $\endgroup$ – djechlin Jan 13 '16 at 1:23
  • $\begingroup$ You can avoid the last square root $4982145>(31·72)^2=4981824$ $\endgroup$ – Jaume Oliver Lafont Feb 9 '16 at 0:13
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Well $\pi^3>3.1415^3>31$. Cubing 3.1415 is not pleasant by hand but most certainly can be done. Or are you looking for a more elegant solution?

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    $\begingroup$ I am looking for a more elegant solution, yes $\endgroup$ – user266519 Jan 12 '16 at 23:18
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    $\begingroup$ Any "elegant" proof is going to involve far more computation than just multiplying twice. $\endgroup$ – DanielV Jan 12 '16 at 23:26
  • $\begingroup$ Clearly the easiest way to prove this is by direct computation. That is why the OP requested a proof without computing the result directly. $\endgroup$ – Elliot G Jan 12 '16 at 23:27
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    $\begingroup$ @DanielV Showing $\pi>3.1415$ requires some ingenuity not to involve a lot of computation. $\endgroup$ – A.S. Jan 13 '16 at 9:00
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It is well known, at least, I know it well :-) that $$\pi>3+\frac1{7+\dfrac1{15}}=3+\frac{15}{106}\ .$$ If you are prepared to accept this, then the rest can certainly be done with $\color{red}{\hbox{mental arithmetic}}$ (or pencil and paper at worst). By the binomial theorem we have $$\pi^3>3^3+3\frac{3^2\times15}{106}+3\frac{3\times15^2}{106^2} =27+\frac{405}{106}+\frac{45^2}{106^2}\ .$$ Now by a familiar trick, $$45^2=2025>1900+120>1900+6\times19=106\times19$$ and so $$\pi^3>27+\frac{405}{106}+\frac{19}{106}=31\ .$$

Comment. The initial inequality is in fact well known to those who know about $\color{red}{\hbox{continued fractions}}$. If it's not familiar in that context, one can quite easily divide mentally to confirm that $\frac{15}{106}<0.14151<\pi-3$.

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The answer is given for this question

Is there an integral for $\pi^4-\frac{2143}{22}$?

Linked to this paper.

http://www.jstor.org/stable/27642693

Extract of what you want:

For $\pi^3$

The truncated continued fractions are $31, 4930/159, 14821/478, \dots$.

\begin{align} \pi^3-31 &= \int_0^1 \frac{8 \, x^5 \, (1-x)^2 \, \left(324889-120736 \, x^2\right)} {445625 \, (1 + x^2) } \log^2 x \, dx\\ \frac{4930}{159}-\pi^3 &= \int_0^1 \frac{4 \, x^{10} \, (1-x)^4 \, \left(695774836+470936528857 \, x^2\right)} {470240754021 \, (1 + x^2) } \log^2 x \, dx. \end{align}

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    $\begingroup$ So we're not allowed to use multiplication (because it wouldn't be "elegant" enough) but we will allow a definite integral involving a ninth-order polynomial and $\log^2$?? Oh so elegant! $\endgroup$ – David G. Stork Jan 12 '16 at 23:32
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    $\begingroup$ Hey, I was only trying to answer the question based on the restrictions... $\endgroup$ – Jack Lam Jan 12 '16 at 23:33
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    $\begingroup$ But Jack Lam's isn't a solution! $\endgroup$ – David G. Stork Jan 12 '16 at 23:39
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    $\begingroup$ @Jack Lam Exactly! That's why this problem should be down voted. $\endgroup$ – David G. Stork Jan 12 '16 at 23:43
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    $\begingroup$ I don't honestly understand why so many people hate this question. Can you please explain? $\endgroup$ – user266519 Jan 12 '16 at 23:45
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From $$\sum_{k=0}^\infty \frac{960}{(2k+1)^6} = \pi^6$$ we have $$\sum_{k=1}^\infty \frac{960}{(2k+1)^6} = \pi^6-960$$ and $$\sum_{k=2}^\infty \frac{960}{(2k+1)^6} = \pi^6-961-\frac{77}{243}$$

Since

$$960<961<961+\frac{77}{243}$$

we can form a new series for $\pi^6-961$ as a weighted sum of the two truncations. Solving the equation

$$\left(\pi^6-960\right)a+\left(\pi^6-961-\frac{77}{243}\right)b=\pi^6-961$$

for rational $a$ and $b$ yields $$a=\frac{77}{320}$$ $$b=\frac{243}{320}$$

Finally, $$\pi^6-961=(\pi^3-31)(\pi^3+31)=3\sum_{k=0}^\infty \left(\frac{77}{(2k+3)^6}+\frac{243}{(2k+5)^6}\right)$$

so

$$\pi^3-31=\frac{3}{\pi^3+31}\sum_{k=0}^\infty \left(\frac{77}{(2k+3)^6}+\frac{243}{(2k+5)^6}\right)$$

is positive because the series contains only positive terms.

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