Re: A Formula for Pi

In article <rubrum-29BD9A.21370621062008@xxxxxxxxxxxxxxxxxxxxx>,
Michael Press <rubrum@xxxxxxxxxxx> wrote:
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>

At <http://www.whim.org/nebula/math/eulerxform.html>, there is a
rigorous proof of the Euler Series Transform.

However, if we apply the acceleration using the identity

--- j C(m,j) k 1
> (-1) -------- = --- ----------
--- C(n+j,k) k+m C(n+m,k+m)
j

proven at <http://www.whim.org/nebula/math/binom.html>, we get

m
--- j C(m,j)
> (-1) ------
--- 1/2+j
j=0

1 1
= --- ------------
m+1 C(1/2+m,m+1)

1 (m+1)!
= --- ---------------
m+1 (1/2+m)...(1/2)

2^{m+1} m!
= ---------------------
(2m+1)(2m-1)...(3)(1)

2^{2m+1} m! m!
= --------------
(2m+1)!

Thus, the Euler Series Transform says

pi

oo
--- j 1
= 4 > (-1) ----
--- 2j+1
j=0

m
--- j 1
= 2 > (-1) -----
--- 1/2+j
j=0

oo m
--- -m-1 --- j C(m,j)
= 2 > 2 > (-1) ------
--- --- 1/2+j
m=0 j=0

oo
--- -m-1 2^{2m+1} m! m!
= 2 > 2 --------------
--- (2m+1)!
m=0

oo
--- 2^{m+1} m! m!
= > -------------
--- (2m+1)!
m=0

Which converges much faster than the Gregory series, better than a
ratio of 2 per term.

Rob Johnson <rob@xxxxxxxxxxxxxx>
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