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Simplifying a sum involving lnx

  • Thread starter dustbin
  • Start date
  • #1
239
5

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.
 

Answers and Replies

  • #2
867
0
ln(a) + ln(b) ≠ ln(a+b), but ln(a) + ln(b) = ln(ab).
Try working with that.
 
  • #3
239
5
Whoops. Actually that was a typo in my first post. I know that ln(a)+ln(b) ≠ ln(a+b):redface:
 
  • #4
3,812
92
Did you try calculating both the limits seperately?
 
  • #5
3,812
92
I have found a solution to this. Continuing from where you left, i can write it as:
[tex]\frac{\ln (n!)-n\ln (n)}{n}[/tex]
which is equal to
[tex]\frac{\ln (\frac{n!}{n^n})}{n}[/tex]

We can write the last term as
[tex]\frac{1}{n}(\ln \frac{1}{n}+\ln \frac{2}{n}+\ln \frac{3}{n}......[/tex]
 
  • #6
239
5
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 :-p

Thank you for your input and help.
 

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