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

n^{n} e^{- n} (2 π n)^{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.