# A series for π

### A series for π

Screenshot_2020-02-29-13-37-48-521_com.bagatrix.mathway.android.jpg (384.9 KiB) Viewed 266 times

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

The first thing I would do is divide both sides by $$\pi- 1$$:
$$\frac{1}{\pi}+ \frac{1}{\pi^2}+ \frac{1}{\pi^3}+ \frac{1}{\pi^4}+ \frac{1}{\pi^5}+ \cdot\cdot\cdot= \frac{1}{\pi- 1}[tex] The sum on the left is the same as [tex]\left(\frac{1}{\pi}\right)+ \left(\frac{1}{\pi}\right)^2+ \left(\frac{1}{\pi}\right)^3+ \left(\frac{1}{\pi}\right)^4+ \left(\frac{1}{\pi}\right)^5+ \cdot\cdot\cdot$$

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 $$\frac{1}{\pi}$$ which has sum $$\frac{1}{1- \frac{1}{\pi}}= \frac{\pi}{\pi- 1}$$. Since there is no "1", subtract 1 from that. The left side of the equation is $$\frac{\pi}{\pi- 1}- 1= \frac{\pi- (\pi- 1)}{\pi- 1}= \frac{1}{\pi- 1}$$.

So the equation is $$\frac{1}{\pi- 1}= \frac{1}{\pi-1}$$.

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

How disappointing!
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