Proving $\pi^3 \gt 31$

Question

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

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?

Answer 1

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.)

Answer 2

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?

Answer 3

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}

Answer 4

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 $\frac{15}{106}<0.14151<\pi-3$.

Answer 5

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.

Answer 6

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.