Re: A Formula for Pi

In article <4f348778-0c41-45e4-bb51-9fcc3dca9916@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Mensanator <mensanator@xxxxxxx> wrote:
On Jun 20, 10:57 am, Maury Barbato <mauriziobarb...@xxxxxxxx> wrote:
Jose Carlos Santos wrote:
On 20-06-2008 7:16, Maury Barbato wrote:

I found in the book "The Penguin Dictionary of
Curious and Interesting Numbers" by Wells the
following formula involving pi

(pi - 3)/4

= sum_{k=1 to infty}[(-1)^(k+1)]/[2k*(2k+1)*(2k+2)]

Is there anybody who knows a proof of this
wonderful
series?

You already got a reply. I only want to remark that
the formula is
equivalent to

pi - 3 = sum_{k = 1 to oo}(-1)^{k + 1}/(k(2k +
2k + 1)(k + 1)

= sum_{k = 1 to oo}(-1)^{k + 1}/(1^2 +
+ 1}/(1^2 + 2^2 + ... + k^2).

Best regards,

Jose Carlos Santos

A little slip. We have

pi - 3 =
sum_{k = 1 to oo}(-1)^{k + 1}/[k(2k+ 1)(k + 1)]
= (1/6)* sum_{k = 1 to oo}(-1)^{k + 1}/(1^2 + 2^2 + ... + k^2)

It's a very very beautiful formula!

Be that as it may, how fast does it converge?
How many terms do I have to sum to get 100 decimal
place accuracy?

He never claimed that it was an efficient way to compute pi, simply
that it was a beautiful formula. It is a personal opinion, but I
agree.

oo
--- k+1 1
> (-1) --------------
--- 2k(2k+1)(2k+2)
k=1

oo
1 --- k-1 1 2 1
= - > (-1) ( -- - ---- + ---- ) [partial fractions]
2 --- 2k 2k+1 2k+2
k=1

oo
1 --- k 1
= - + > (-1) ---- [collapse telescoping terms]
4 --- 2k+1
k=1

1 pi
= - + ( -- - 1 ) [Gregory series]
4 4

pi - 3
= ------
4

Being an alternating series with monotonically decreasing terms, the
error is less than 1/(8n^3) after n terms.

Rob Johnson <rob@xxxxxxxxxxxxxx>
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Relevant Pages

  • Re: A Formula for Pi
    ... Mensanator wrote: ... Curious and Interesting Numbers" by Wells the ... that it was a beautiful formula. ...     oo ...
    (sci.math)
  • Re: A Formula for Pi
    ... Mensanator wrote: ... that it was a beautiful formula. ...     oo ... Being an alternating series with monotonically decreasing terms, ...
    (sci.math)