Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Increasing/decreasing sequences

  1. May 17, 2011 #1
    I have been asked to find if the following sequence is increasing or decreasing:
    an = ne^-n

    So I first multiplied thru by e^n to get: n/e^n
    Then, I did n+1, so (n+1)*e^-(n+1). I moved the negative exponent to the bottom to get
    (n+1)/(e^(n+1))

    I guess my first question is did I start off correctly.
     
  2. jcsd
  3. May 17, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    Insights Author

    Yes, so the question is, which of the following two holds:

    [tex]\frac{n}{e^n}\leq \frac{n+1}{e^{n+1}}~\text{or}~\frac{n}{e^n}\geq \frac{n+1}{e^{n+1}}[/tex]

    Try to simplify these expressions...
     
  4. May 17, 2011 #3

    Mark44

    User Avatar
    Insights Author

    Staff: Mentor

    What you have is equal to ne-n, but you didn't get it by multiplying by en, which would have been wrong to do.

    e-n is equal to 1/en. What you did was rewrite the original expression in different form, using a property of exponents.
    ???
    You're working with an expression, ne-n. Unlike an equation, there are only a few things you can do with an equation. The only number you can add to an expression is 0 - otherwise you will change the value of the expression. The only number you can multiply by is 1 - which can take many forms, but it's still 1. If you multiply by something other than 1, you will get a new expression that is not equal to the one you started with.

    So in particular, you can't just multiply by n + 1, and you can't just decide to change the exponent from -n to -(n + 1). Those are not valid operations here.
     
  5. May 17, 2011 #4
    I don't know if you are allowed to do this, but the standard way is to take the derivative of f(x)=xe-x and show that it's negative for x>1, thus is a decreasing function on that interval. Since an=f(n), this gives that the sequence is decreasing.
     
  6. May 17, 2011 #5

    Mark44

    User Avatar
    Insights Author

    Staff: Mentor

    This is one way. micromass is showing another way that more directly shows that the sequence is decreasing.
     
  7. May 17, 2011 #6
    Right, which means you have to start with one side, and work to the other. It's not clear how to do that directly. Technically, you cannot start with either of those assumptions and simplify, though doing that might give you a hint with how to work in a more direct manner.
     
  8. May 17, 2011 #7

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    Insights Author

    In this particular case, my approach is not that hard. Just bring the n+1 and the en to the other side, and you'll see something nice popping up!

    I do agree that your approach would be best in almost every other case, as derivatives are quite easy to handle :smile: In fact, I would have mentioned the derivative thingy, but I forgot about it :frown:
     
  9. May 17, 2011 #8

    Mark44

    User Avatar
    Insights Author

    Staff: Mentor

    Not necessarily.
    You can start with this expression,
    [tex]\frac{n}{e^n} - \frac{n+1}{e^{n+1}}[/tex]
    and show that it is positive. That shows that an > an+ 1, or equivalently, that the sequence is decreasing.
     
  10. May 17, 2011 #9
    RIGHT, forgot about doing that. Thanks!
     
  11. May 17, 2011 #10
    And I forgot about the subtraction technique!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Increasing/decreasing sequences
  1. Increasing Sequence (Replies: 1)

Loading...