Re: could someone help me identify a number?



Le 14/05/11 12:21, Franz Gnaedinger a écrit :
Could someone help me identify a number,
the limit of the series

1/(1x4) + 1/(7x10) + 1/(13x16) ... ?

In my simpler notation

'1x4 '7x10 '13x16 '19x22 '25x28 ...

It is about 0.278497.



S= sum_{n=0}^oo 1/((6*n+1)*(6*n+3)) = ln(3)/8 + pi*sqrt(3)/24

A general method to find this kind of sums is explained here : <http://people.math.sfu.ca/~cbm/aands/page_264.htm>

Rewrite S as :
S= 1/(2*6) sum_{n=1}^oo 1/(n-5/6) - 1/(n-1/2)
= (-psi(1-5/6) + psi(1-1/2))/12

with psi the digamma function admitting following formula :

<http://en.wikipedia.org/wiki/Digamma_function#Gauss.27s_digamma_theorem>

Hoping this helped,
Raymond


Thanks in advance.

'1x2 '2x3 '3x4 '4x5 '5x6 ... = 1
'1x2 '3x4 '5x6 '7x8 '9x10 ... = ln2

'2 plus '(6x6 + 2) plus '1x2x3
'2 plus '(10x10 + 2) plus '1x2x3 '3x4x5
'2 plus '(14x14 + 2) plus '1x2x3 '3x4x5 '5x6x7

and so on

'1x3 '3x5 '5x7 '7x9 '9x11 ... = '2
'1x3 '5x7 '9x11 '13x15 '17x19 ... = pi/8

'4 plus '(10x10 + 8) plus 2 times '1x3x5
'4 plus '(18x18 + 8) plus 2 times '1x3x5 '5x7x9
'4 plus '(26x26 + 8) plus 2 times '1x3x5 '5x7x11 '11x13x15

and so on

'1x4 '4x7 '7x10 '10x13 '13x16 ... = '3
'1x4 '7x10 '13x16 '19x22 '25x28 ... ??? about 0.278497

'6 plus '(14x14 + 17) + 3 times '1x4x7
'6 plus '(26x26 + 17) + 3 times '1x4x7 '7x10x13
'6 plus '(38x38 + 17) + 3 times '1x4x7 '7x10x13 '13x15x17

and so on

'1x5 '5x9 '9x13 '13x17 '17x21 ... = '4
'1x5 '9x13 '17x21 '25x29 '33x37 ... = ???? (0.216...)

'8 plus '(18x18 + 29) + 4 times '1x5x9
'8 plus '(34x34 + 17) + 4 times '1x5x9 '9x13x17
'8 plus '(50x50 + 17) + 4 times '1x5x9 '9x13x17 '17x21x25

and so on

.



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