Fun ways to find the bounds of π2\pi^2 using Riemann Zeta Functions!

We all know that n=11n(n+1)=n=1(1n1n+1)=1 \displaystyle \sum_{n=1}^\infty \frac1{n(n+1)} =\sum_{n=1}^\infty \left(\frac1n -\frac1{n+1}\right) = 1 by partial fractions followed by telescoping sum. But if we raise the expression 1n(n+1)\frac1{n(n+1)} to the power of any positive number greater than 1, something interesting happens!

By partial fractions, (1n(n+1))3=(1n31(n+1)3)(1n2+1(n+1)2)6(1n1n+1) \left(\frac1{n(n+1)}\right)^3 = \left(\frac1{n^3} - \frac1{(n+1)^3}\right) - \left(\frac1{n^2} + \frac1{(n+1)^2} \right) - 6 \left(\frac1n - \frac1{n+1} \right) .

With the knowledge of the Basel function: n=11n2=π26\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6 .

n=1(1n(n+1))3=n=1[(1n31(n+1)3)3(1n2+1(n+1)2)+6(1n1n+1)]=(1)3(2(π26)1)+6(1)=10π2 \begin{aligned} &&\displaystyle \sum_{n=1}^\infty \left(\frac1{n(n+1)}\right)^3 \\ &=& \displaystyle \sum_{n=1}^\infty \left[ \left(\frac1{n^3} - \frac1{(n+1)^3}\right) - 3\left(\frac1{n^2} + \frac1{(n+1)^2} \right) + 6 \left(\frac1n - \frac1{n+1} \right) \right ] \\ &=& \displaystyle (1) - 3 \left( 2\left( \frac{\pi^2}{6}\right) - 1\right) + 6(1) \\ &=& 10 - \pi^2 \\ \end{aligned}

Because n=1(1n(n+1))3 \displaystyle \sum_{n=1}^\infty \left(\frac1{n(n+1)}\right)^3 is the sum of all positive terms, then it must be strictly positive, then 10π2>010 - \pi^2 > 0 , equivalently π2<10\pi^2 < 10 . Thus we just found an upper bound of π2\pi^2. Wonderful isn't it?

You could keep increasing the value of mm for the expression (1n(n+1))m \left(\frac1{n(n+1)}\right)^m for positive number mm to increase the accuracy of the bound!

Try it yourself!

Question 1: Prove the partial fraction (1n(n+1))2(1n2+1(n+1)2)2(1n1n+1) \left ( \frac1{n(n+1)} \right)^2 \equiv \left ( \frac 1{n^2} + \frac1{(n+1)^2} \right) - 2 \left (\frac1n -\frac1{n+1}\right). And determine the value of n=1(1n(n+1))2 \displaystyle \sum_{n=1}^\infty \left ( \frac1{n(n+1)} \right)^2 . Thus show that π>3\pi > 3 .

Question 2: Given that n=11n4=π490\displaystyle \sum_{n=1}^\infty \frac1{n^4} = \frac{\pi^4}{90} and find the partial fraction of (1n(n+1))5 \left(\frac1{n(n+1)}\right)^5 , prove that π2<32(172935)\pi^2 < \frac32 \left(\sqrt{1729}-35\right) . Consider the example given and the answer you've found, can you determine a systematic way to increase the upper bound of π2\pi^2 ?

Question 3: Suppose we restrict mm (as mentioned above) to an odd number, why does the accuracy of the upper bound increases when the number mm increases?

Question 4: Consider n=1(1n(n+1)(n+2))2 \displaystyle \sum_{n=1}^\infty \left( \frac1{n(n+1)(n+2)}\right)^2 , prove that π>392\pi> \frac{ \sqrt{39}}2 . Similarly, consider n=1(1n(n+1)(n+2)(n+3))2 \displaystyle \sum_{n=1}^\infty \left( \frac1{n(n+1)(n+2)(n+3)}\right)^2 , prove that 9.85<π2<109.85 < \pi^2 < 10.

Question 5 Using the answers you got from Question 2, find the partial fractions for (1n(n+1))4 \left(\frac1{n(n+1)}\right)^4 , and thus prove the inequality 15(425)<π2<32(172935). \large 15(4\sqrt2 -5) < \pi^2 < \frac32 \left(\sqrt{1729}-35\right).

Question 6 (warning: tedious): Like in Question 4, consider n=1(1n(n+1)(n+2)(n+3)(n+4))2 \displaystyle \sum_{n=1}^\infty \left( \frac1{n(n+1)(n+2)(n+3)(n+4)}\right)^2 , prove that π2>1105112 \pi^2 > \frac{1105}{112} .

Question 7: From Question 4 and 6, can you spot a pattern to find a lower bound of π2\pi^2 ? Prove that using this technique you've found, you can always find a rational number AA such that π2>A\pi^2 > A .

Question 8 (warning: very tedious): If we are further given that ζ(6)=n=11n6=π6945\displaystyle \zeta(6) = \sum_{n=1}^\infty \frac1{n^6} = \frac{\pi^6}{945} , and find the partial fraction of (1n(n+1))7 \left(\frac1{n(n+1)}\right)^7 and by Cardano's method, prove that π2<42+3(7486+6279146531897486+627914653) \large \pi^2 < -42 + 3 \left ( \sqrt[3]{7486 + \sqrt{62791465}} - \frac{189}{\sqrt[3]{7486 + \sqrt{62791465}}} \right ) .

Question 9 (warning: extremely tedious): From Question 8, if we are again further given that ζ(8)=π89450\zeta(8) = \frac{\pi^8}{9450}, using the approaches we had before, prove that the equation below has a positive real root BB such that Bπ2B - \pi^2 is a significantly small positive number.

3x4+550x3+45045x2+3378375x38288250=0\large 3x^4 + 550x^3 + 45045x^2 + 3378375x - 38288250 = 0

Note by Pi Han Goh
4 years, 5 months ago

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If we continue to use this trick, will it eventually converge to pi ( or pi^2,pi^4 etc), or will it always be really close, but not pi itself?I doubt it, because every time we have an algebraic number as an approximation and pi is transcendental.

Bogdan Simeonov - 4 years, 5 months ago

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Woah! I'll check this out soon :)

Jake Lai - 4 years, 5 months ago

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I love it :) .........Sir finding an estimate for pi requires the zeta at even integers ..... can we reverse this method to instead approximate for zeta at odd integers?

Abhinav Raichur - 4 years, 5 months ago

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Cool! Please keep on sharing such awesome things like this with us .

A Former Brilliant Member - 4 years, 5 months ago

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Just asking, is the polynomial in question 88 wrong? I'm not sure about it but I graphed it out and there was no root near π2\pi^{2}. It may be possible that Desmos screwed up because the numbers are too big.

Julian Poon - 4 years, 5 months ago

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LINK?

Pi Han Goh - 4 years, 5 months ago

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It's correct now... I think I made a mistake typing it last night.

Julian Poon - 4 years, 5 months ago

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