Simplifying a sum involving lnx

[dustbin]

Homework Statement

I need to show that the lim n->inf ( [(1/n) * sum from k=1 to n of ln(k) ] - ln(n) )
is equal to the lim n->inf ( (1/n) sum from k=1 to n of ln(k/n) )

The Attempt at a Solution

I showed that the sum of ln(n) from k=1 to n is ln(n!) using ln(a) + ln(b) = ln(a+b). I am not sure how to get from (1/n)ln(n!) - ln(n) to (1/n)sum-ln(k/n).


I apologize for the non-tex stuff... but I am not sure how to do some of the symbols for this in tex yet. Thank you for your help.

 

[Bohrok]

ln(a) + ln(b) ≠ ln(a+b), but ln(a) + ln(b) = ln(ab).
Try working with that.

 

[dustbin]

Whoops. Actually that was a typo in my first post. I know that ln(a)+ln(b) ≠ ln(a+b)

 

[Saitama]

Did you try calculating both the limits seperately?

 

[Saitama]

I have found a solution to this. Continuing from where you left, i can write it as:
\frac{\ln (n!)-n\ln (n)}{n}
which is equal to
\frac{\ln (\frac{n!}{n^n})}{n}

We can write the last term as
\frac{1}{n}(\ln \frac{1}{n}+\ln \frac{2}{n}+\ln \frac{3}{n}...

 

[dustbin]

Oooh. I got up to (1/n)ln(n!/n^n) but did not see that it gives what you have for the last step. After thinking about it I can see why this is. Also, I did try to calculate both limits separately... but I could not find a way to the solution by doing so. I appreciate your input.

I was looking at ln(n!) as [ ln(n) + ln(n-1) + ln(n-2) + ... + ln(2) + ln(1) ] instead of in the opposite order, or [ ln(1) + ln(2) + ln(3) + ... ln(n-1) + ln(n) ]. After I reversed the order, I saw how you obtained the solution

Thank you for your input and help.