closed form solution for summation of $\log(i)$

Question

Is there a way to find a closed form solution for: (Note that base is $2$)

$\displaystyle\sum_{i=1}^n\log_2(i)$

thanks for any help Can't find a formula for this

Could we please change the index of summation here to something besides "$i$"?
@ChrisTaylor: It makes the question title ambiguous and confusing. Is the question about the power series for the logarithm of $i=\sqrt{-1}$? (which is what I thought when I first saw the question title)

Myself · Accepted Answer · 2012-02-01 06:10:29Z

up vote 4 down vote accepted

$$ \sum_{i=1}^n \log i = \log \left(\prod_{i=1}^n i\right) = \log (n!).$$

But note that the left hand site is actually easier to compute (numerically).

edited Feb 1 '12 at 6:10

answered Feb 1 '12 at 5:51

Myself
4,1411025

2

Do you mean log(n!) at the end? – cssl Feb 1 '12 at 6:03

@cssk, Oops, I do, thanks! – Myself Feb 1 '12 at 6:10

1

Note that you can obtain nice asymptotics for the whole expression by Stirling's formula. – Raphael Feb 1 '12 at 6:20

1 Answer