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.

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

Woah! I'll check this out soon :)

Log in to reply

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?

Log in to reply

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

Log in to reply

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.

Log in to reply

LINK?

Log in to reply

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

Log in to reply