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

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

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

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

$\begin{eqnarray} &&\displaystyle \sum_{n=1}^\infty \left(\frac1{n(n+1)}\right)^3 \\ &=& \displaystyle \sum_{n=1}^\infty \left[ \left(\frac1{n^3} - \frac1{(n+1)^3}\right) - 3\left(\frac1{n^2} + \frac1{(n+1)^2} \right) + 6 \left(\frac1n - \frac1{n+1} \right) \right ] \\ &=& \displaystyle (1) - 3 \left( 2\left( \frac{\pi^2}{6}\right) - 1\right) + 6(1) \\ &=& 10 - \pi^2 \\ \end{eqnarray}$

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

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

Try it yourself!

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

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

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

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

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

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

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

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

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

$\large 3x^4 + 550x^3 + 45045x^2 + 3378375x - 38288250 = 0$

Note by Pi Han Goh
1 year, 8 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. · 1 year, 8 months ago

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Just asking, is the polynomial in question $$8$$ wrong? I'm not sure about it but I graphed it out and there was no root near $$\pi^{2}$$. It may be possible that Desmos screwed up because the numbers are too big. · 1 year, 8 months ago

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LINK? · 1 year, 8 months ago

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It's correct now... I think I made a mistake typing it last night. · 1 year, 8 months ago

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Cool! Please keep on sharing such awesome things like this with us . · 1 year, 8 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? · 1 year, 8 months ago

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Woah! I'll check this out soon :) · 1 year, 8 months ago

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