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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

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Note that $\log a+\log b = \log (ab)$ –  E.O. Feb 1 '12 at 6:07
    
Could we please change the index of summation here to something besides "$i$"? –  deoxygerbe Feb 1 '12 at 6:30
    
@deoxygerbe What's wrong with $i$? –  Chris Taylor Feb 1 '12 at 7:52
    
@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) –  deoxygerbe Feb 1 '12 at 7:58
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1 Answer 1

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).

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