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I'm trying to simplify the following summation: $$\sum_{k=1}^{n-1} (2k + \log_2(k) - 1)$$.

I've basically done the following:

$$\sum_{k=1}^{n-1} (2k + \log_2(k) - 1) \\ = \sum_{k=1}^{n-1} 2k + \sum_{k=1}^{n-1} \log_2(k) - \sum_{k=1}^{n-1} 1\\ = \frac{n(n-1)}{2} + \sum_{k=1}^{n-1} \log_2(k) - (n-1)$$

Now I'm trying to do deal with this term $\sum_{k=1}^{n-1} \log_2(k)$, but I'm a bit confused.

My gut tells me I can do the following:

$$\sum_{k=1}^{n-1} \log_2(k)\\ = \log_2(1) + \log_2(2) + \ldots + \log_2(n-1)\\ = \log_2(\prod_{k=1}^{n-1} k)\\ = \log_2((n-1)!)$$

Using that $\log_a(b) + \log_a(c) = \log_a(b \cdot c)$.

However, I'm not convinced this is an entirely valid reasoning because I can't find any rules/identities for dealing with $\sum_{k=1}^{n-1} \log_2(k)$. Is this correct or are there any rules to apply?

J. M. ain't a mathematician
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Mythio
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    Your reasoning is perfectly valid, and the validity of the simplification for any number of log terms follows easily from the fact that $\log x + \log y = \log(xy)$. – Old John Apr 11 '13 at 11:04
  • Ok good. Does that mean the rule $\sum_{k=1}^{n} \log_a(k) = \log_a(\prod_{k=1}^{n} k)$ holds in general? – Mythio Apr 11 '13 at 11:08
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    Shouldn't it be $n(n-1)$ instad of $\frac{n(n-1)}{2}$ since you do the summation of $\sum_{k=1}^{n-1} 2k$? – Thomas Apr 11 '13 at 11:09
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    @Mythio yes it does! You already stated that $\log(b) +\log(c) = \log(b\cdot c)$. Now you can simply add further terms and you get the rule you mention. – Thomas Apr 11 '13 at 11:10
  • @macydanim You are absolutely right, I forgot the 2 infront of the summation (I always move the multiplication constant out of the summation first). However, the end result would indeed be $n(n-1)$ EDIT: sigh...got to keep paying attention.. – Mythio Apr 11 '13 at 11:10
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    @Mythio sorry I don't understand that. If you have $\sum_{k=1}^{n-1} 2k$ then the result is $n(n-1)$. I don't know what you state with "However, the end result would ..." – Thomas Apr 11 '13 at 11:12
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    @macydanim: third time is a charm...its always in the details – Mythio Apr 11 '13 at 11:15

1 Answers1

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(Just so people know this has been answered)

You right

$$\sum_{k=1}^{n} \log k = \log (n!)$$

If you want to justify it formally, you can try using induction.

Aryabhata
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