This question is motivated by Why is $\pi$ so close to $3$?, Why is $\pi^2$ so close to $10$? and Proving $\pi^3 \gt 31$.

I. $\pi$ and $\pi^2$

There are series with all terms positive for $\pi-3$ and $10-\pi^2$


(see Lehmer, http://matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27121.pdf page 139) and, $$10-\pi^2=\sum_{k=0}^\infty\frac{1}{((k+1)(k+2))^3}$$

II. $\pi^3$

Regarding $\pi^3\approx 31$, one may equivalently express $\pi^6\approx 961$ from Euler's series

$$\frac{\pi^6}{945}=\sum_{k=0}^\infty \frac{1}{(k+1)^6}$$ and (see Proving $\pi^3 \gt 31$), $$\frac{\pi^6}{960}-1=\sum_{k=0}^\infty \frac{1}{(2k+3)^6}$$

Multiplying the first by $\displaystyle-\frac{63}{2^6}$ and the second by $31^2=961$, then adding them,

$$\pi^6-961=\sum_{k=0}^\infty\left(-\frac{63}{(2k+2)^6}+\frac{961}{(2k+3)^6}\right)$$ This series can be rewritten in a single positive fraction per term as $$\pi^6-961=\sum_{k=0}^\infty\frac{57472 k^6+332736 k^5+786480 k^4+957920 k^3+616380 k^2+185316 k+15577}{64 (k+1)^6 (2 k+3)^6}$$

so it directly proves $\pi^6>961$ and hence $\pi^3>31$. However, this lacks the simplicity of the ones for $\pi-3$ and $10-\pi^2$.

Q: How can we write $\pi^6−961$ as $\sum_{k=0}^{\infty} \frac{1}{P(k)}$ where $P$ is a polynomial with nonnegative, rational coefficients?

[EDIT] A direct series to prove $\pi^3>31$ is given by

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


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    $\begingroup$ This question is unfocused, and rather hard to read IMO (or more like, doesn't really make you wanna read it). Nevertheless, there seems to be quite a lot of effort put into it, so I don't understand why it should be down-voted, in particularly, without leaving any comment explaining it. $\endgroup$ Feb 9, 2016 at 11:12
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    $\begingroup$ @GerryMyerson A reasonable and natural question in this context is : can we write $\pi^6-961$ as a sum of the form $\sum_{k=1}^{n} \frac{1}{P(k)}$ where $P$ is a polynomial with nonnegative, rational coefficients. This is the case for the examples given in the OP $\endgroup$ Feb 9, 2016 at 13:01
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    $\begingroup$ @NoChance That has already been used: when combined with the shortest zero relation it gives the formula for $\pi-3$. math.stackexchange.com/a/1639961/134791 The problem here is $\pi^6-961$ or $\pi^3-31$. $\endgroup$ Feb 9, 2016 at 22:22
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    $\begingroup$ @barakmanos: I think I've improved his post a bit. It's quite interesting, IMO. $\endgroup$ Feb 10, 2016 at 16:55
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    $\begingroup$ From one single series, $$\pi^6-961=3\sum_{k=0}^\infty \left(\frac{77}{(2k+3)^6}+\frac{243}{(2k+5)^6}\right)$$ math.stackexchange.com/a/1651175/134791 $\endgroup$ Feb 13, 2016 at 23:25

1 Answer 1


I try to find the following formula from another perspective, and it is easier to generalize.

$$ \pi^6-961=\sum_{k=0}^\infty \left(\frac{231}{(2k+3)^6}+\frac{729}{(2k+5)^6}\right) $$


$$ \sum _{k=0}^{\infty } \left(\frac{a}{(2 k+1)^6}+\frac{b}{(2 k+3)^6}+\frac{c}{(2 k+5)^6}\right) $$

The exponent of the denominator must be $6$, otherwise $π^6$ cannot be obtained.

The internal form is $k + b$ or $2 k + b$. Other forms are very complicated.

Summing up we get:

$$ \frac{\pi ^6 }{960}a+\frac{1}{960} \left(\pi ^6-960\right) b+\left(\frac{\pi ^6}{960}-\frac{730}{729}\right) c=\pi ^6-961 $$

Take common division and rearrangement are:

$$ \pi ^6 (243 a+243 b+243 c-233280)-233280 b-233600 c+224182080 = 0 $$

Note that $a$ appears alone to the power of $6$, so $a$ must be $0$.

The rest is just a matter of solving a simple linear Diophantine equation:

$$ \begin{cases} 233280 b+233600 c-224182080=0\\ 243 b+243 c-233280=0\\ a⩾ 0,b⩾0,c⩾0 \end{cases} $$

The only solution is:

$$ \begin{cases} a = 0 \\ b = 231\\ c = 729\\ \end{cases} $$


$$ \pi^6-961=\sum_{k=0}^\infty \left(\frac{231}{(2k+3)^6}+\frac{729}{(2k+5)^6}\right) > 0 $$


$$ \pi^3 > \sqrt{961} = 31 $$

Fortunately, we can get the solution directly. If we don't get the solution, we can get other forms of solutions by adjusting the denominator and the number of terms.

For example:

$$ \pi^6-961=\sum _{k=0}^{\infty } \left(\frac{2889}{(2 k+6)^6}+\frac{57591}{(2 k+8)^6}\right) $$


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