A series for π

A series for π

Postby Guest » Sat Feb 29, 2020 3:47 pm

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Does anyone know if there is any way to approximate π using this formula?
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Re: A series for π

Postby Guest » Sun Apr 05, 2020 9:14 am

The first thing I would do is divide both sides by [tex]\pi- 1[/tex]:
[tex]\frac{1}{\pi}+ \frac{1}{\pi^2}+ \frac{1}{\pi^3}+ \frac{1}{\pi^4}+ \frac{1}{\pi^5}+ \cdot\cdot\cdot= \frac{1}{\pi- 1}[tex]

The sum on the left is the same as
[tex]\left(\frac{1}{\pi}\right)+ \left(\frac{1}{\pi}\right)^2+ \left(\frac{1}{\pi}\right)^3+ \left(\frac{1}{\pi}\right)^4+ \left(\frac{1}{\pi}\right)^5+ \cdot\cdot\cdot[/tex]

If there were a "1" at the beginning of the sum on the left, it would be a geometric sum with first term 1 and constant multiple [tex]\frac{1}{\pi}[/tex] which has sum [tex]\frac{1}{1- \frac{1}{\pi}}= \frac{\pi}{\pi- 1}[/tex]. Since there is no "1", subtract 1 from that. The left side of the equation is [tex]\frac{\pi}{\pi- 1}- 1= \frac{\pi- (\pi- 1)}{\pi- 1}= \frac{1}{\pi- 1}[/tex].

So the equation is [tex]\frac{1}{\pi- 1}= \frac{1}{\pi-1}[/tex].

That is an "identity" true for any value of [tex]\pi[/tex]! So, no, you cannot calculate a value of [tex]\pi[/tex] from that formula.


How disappointing!
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