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Homework Help: Maximum value of a_n

  1. Jan 29, 2013 #1
    1. The problem statement, all variables and given/known data
    (These may not be the exact wordings as it was asked by my friend and I do not have the exact question)
    Find the value of n for which a_n is maximum where
    [tex]a_n=\frac{1000^n}{n!}[/tex]

    (Ans: n=999)

    2. Relevant equations



    3. The attempt at a solution
    I don't think using calculus in this type of question would help so I have put it in Precalc section. I am clueless here, how should I begin with this.
     
  2. jcsd
  3. Jan 29, 2013 #2

    CompuChip

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    You could try looking at [tex]\frac{a_{n + 1}}{a_n}[/tex]
     
  4. Jan 29, 2013 #3

    rollingstein

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    I took log of both sides and then used the Stirling Approximation.

    Then differentiate with respect to n and set that to zero.

    I get n=1000 (but there must be some ugliness because of the discontinuity etc.)
     
  5. Jan 29, 2013 #4

    CompuChip

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    If you follow my hint, you will find a cleaner approach.
     
  6. Jan 29, 2013 #5
    This is equal to ##\frac{1000}{n+1}##.
    But I am still clueless.
     
  7. Jan 29, 2013 #6

    rollingstein

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    If the series in increasing what's the value of the ratio. If it is decreasing what's the value of the ratio?

    @CompuChip: Yes, that's cleaner.
     
  8. Jan 29, 2013 #7

    rollingstein

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    In any case shouldn't 999 and 1000 both be acceptable answers?
     
  9. Jan 29, 2013 #8
    If the series increases,its greater than 1 and less than 1 if it decreases.
     
  10. Jan 29, 2013 #9

    rollingstein

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    You still don't see the answer?
     
  11. Jan 29, 2013 #10
    Honestly, no.
     
  12. Jan 29, 2013 #11

    CompuChip

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    So what happens at the maximum?
     
  13. Jan 29, 2013 #12
    Can you tell me to which area of mathematics these type of questions belong?
     
  14. Jan 29, 2013 #13
    @CompuChip: Sorry if this is annoying you but I really have no idea.
     
  15. Jan 29, 2013 #14

    CompuChip

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    Look at the bit that I quoted: you say that if the series is increasing, then the ratio between consecutive elements is > 1. If it is decreasing, then the ratio is < 1.

    Now imagine a series with a maximum. Draw one on a piece of paper, if it helps. What can you say about the increasing / decreasing on either side of the maximum? What does that tell you about the ratio "at" the maximum?
     
  16. Jan 29, 2013 #15
    Before the maximum, series increases. After that it decreases.
     
  17. Jan 29, 2013 #16

    CompuChip

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    So before the maximum [itex]a_{n+1}/a_n > 1[/itex] and after that [itex]a_{n+1}/a_n < 1[/itex]. What happens in between?
     
  18. Jan 29, 2013 #17

    SammyS

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    Well it is pretty easy to show that a999 = a1000 .
     
  19. Jan 29, 2013 #18
    But wait!

    [itex]\frac{1000^{999.5}}{999.5!}[/itex]

    is larger than either of those.....

    And now we go to scary places.
     
  20. Jan 29, 2013 #19
    Equal to 1. Therefore n=999.

    I plugged a_n in wolframalpha with n=999 and n=1000. The results are same but how can I show that? :confused:

    As I do not have the exact question, it should be mentioned that n is an integer.
     
  21. Jan 29, 2013 #20

    Dick

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    n=999 gives 1000^999/999!. n=1000 gives 1000^1000/1000!. 1000!=999!*1000.
     
  22. Jan 29, 2013 #21
    Thanks Dick! :smile:
     
  23. Jan 29, 2013 #22
    I still want to ask one question. How you came up with this or how did this pop up in your mind? I am interested to know how you think about these type of problems. :smile:
     
  24. Jan 30, 2013 #23

    ehild

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    Having a sequence of positive numbers, if nth is the greatest among them, it must be greater then its neighbours, so an+1≤an and an-1≤an:biggrin:

    ehild
     
  25. Jan 30, 2013 #24

    Ray Vickson

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    You already have all you need: you have a_1000/a_999 = 1, so a_999 = a_1000 !
     
  26. Jan 30, 2013 #25

    CompuChip

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    You had already shown that
    [tex]\frac{a_{n+1}}{a_n} = \frac{1000}{n + 1}[/tex]
    If you set that equal to 1, you can solve for n quite straightforwardly.

    I guess this is one of those things where a bit of experience helps... I was initially looking at the limit for [itex]n \to \infty[/itex] and thought of the ratio test. Then I noticed that in the ratio the factorial disappeared, and somehow made the connection to your actual question.
     
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