Re: A Formula for Pi

On 20-06-2008 18:15, Mensanator 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?

We are dealing with an alternating series here. Now, for such a series

a_1 - a_2 + a_3 - a_4 + ...

if s_n is the sum if its _n_ first terms and if _s_ is he sum of the
series, then |s - s_n| < a_{n + 1}. So, take _n_ such that
1/(n(n + 1)(2n + 1)) < 10_{-101} <=> n(n + 1)(2n + 1) > 10^{101}. Taking
n = 10^{34} is enough.

So, yes, it converges slowly.

Best regards,

Jose Carlos Santos
.

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