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.

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

## Comments

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TopNewestIf 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 · 2 years, 4 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. – Julian Poon · 2 years, 4 months ago

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– Pi Han Goh · 2 years, 4 months ago

LINK?Log in to reply

– Julian Poon · 2 years, 4 months ago

It's correct now... I think I made a mistake typing it last night.Log in to reply

Cool! Please keep on sharing such awesome things like this with us . – Azhaghu Roopesh M · 2 years, 4 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 · 2 years, 4 months ago

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Woah! I'll check this out soon :) – Jake Lai · 2 years, 4 months ago

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