Re: A Formula for Pi

In article
<41c3a2f3-d399-42f6-b21d-ce897b6ba6c5@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Mensanator <mensanator@xxxxxxx> wrote:

On Jun 20, 2:35 pm, r...@xxxxxxxxxxxxxx (Rob Johnson) wrote:
In article <4f348778-0c41-45e4-bb51-9fcc3dca9...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,





Mensanator <mensana...@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,

I didn't say he claimed it was efficient. I'm _asking_
whether or not it is efficient.

simply
that it was a beautiful formula.  

Fine. It's beautiful. No disagreemrnt there.

It is a personal opinion,

Which everyone is entitled to.

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.

Some of us don't know how to do this. But given the series, I can
write a program to apply it.

But I would prefer a series that converges in ~300 terms to the
one that converges in ~10**34 terms.

How about Ramanujan's series that gets eight decimal places for each term?

1 2.sqrt{2} (4n)! [1103 + 26390n]
-- = --------- sum_n ------- ---------------
pi 9801 (n!)^4 (4*99)^4n

Also look up Euler's (who else?) transformation for accelerating
alternating series. In essence a difference table.
Abramowitz and Stegun, 3.6.27.

<http://mathworld.wolfram.com/EulersSeriesTransformation.html>

--
Michael Press
.

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: ... Curious and Interesting Numbers" by Wells the ... A little slip. ... It's a very very beautiful formula! ...
    (sci.math)
  • Re: "Musatov"...
    ... disrespectful. ...    And let me assure you, ... force, Mensanator. ...
    (sci.math)