log(1)+log(2)+log(3)+...
By Chi Kin Ho (P1942) on Monday, August
7, 2000 - 07:43 pm :
Can anyone tell me whether this is a series:
log(1)+log(2)+log(3)+ ... ?
If so, how do you calculate the sum to a particular term, because
I don't seem to find a common ratio or difference?
Thanks,
Chi
By Dan Goodman (Dfmg2) on Tuesday,
August 8, 2000 - 07:14 am :
It is certainly a series, but it's not
an arithmetic or geometric series. I don't think you can do any
better than log(1)+log(2)+...+log(n)=log(n!), where
n!=n(n-1)(n-2)...(3)(2)(1) (n factorial).
By Anonymous on Monday, August 14, 2000
- 02:41 am :
Chi:
An approximate answer to the sum to nth term can be found using
the Stirling's formula for n!.
By Dave Sheridan (Dms22) on Friday,
August 25, 2000 - 10:19 am :
Stirling's formula is that n! can be approximated very
well by nn e-n(2pn)1/2. Actually, Stirling's theorem is that n!
divided by this converges to 1 as n tends to infinity. The main use is that
it converts factorials to more easily manipulated expressions with a tiny error.
Try seeing the difference between the two expressions for n=2, 6, 10, 14. By
the last of those, I think the match is very close indeed.
-Dave
By Thomas Mooney (P3048) on Friday,
October 20, 2000 - 08:34 pm :
You could go to
SOS maths to get a derivation of the formula, but be careful
as the proof is very rigorous and some parts might be unclear.