1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Having trouble manipulating the equation

  1. Oct 22, 2006 #1

    kreil

    User Avatar
    Gold Member

    Using the fact that [tex]E_n=(1+\frac{1}{n})^n {\rightarrow}e[/tex], find the limits of the following:

    a) [tex](1+\frac{1}{n^2})^{n^2}[/tex]

    b)[tex](1+\frac{1}{2n})^n[/tex]

    c)[tex](1+\frac{2}{n})^n[/tex]


    Here is the work I've done:

    a) If u=n2, [tex](1+\frac{1}{n^2})^{n^2}=(1+\frac{1}{u})^u {\rightarrow} e[/tex]

    b) [tex](1+\frac{1}{2n})^n=((1+\frac{1}{2n})^{2n})^{\frac{1}{2}} {\rightarrow} e^{\frac{1}{2}}[/tex]

    c) This is the part I'm having problems with. I'm having trouble manipulating the equation so that I can use the given fact about convergence to e. Any suggestions will be helpful.

    Josh
     
    Last edited: Oct 22, 2006
  2. jcsd
  3. Oct 22, 2006 #2

    StatusX

    User Avatar
    Homework Helper

    It's the same as b except with 1/2 instead of 2.
     
  4. Oct 22, 2006 #3

    kreil

    User Avatar
    Gold Member

    Ohh ok, got that now.

    Another question I'm having problems with is that I need to come up with a bounded sequence that is not Cauchy. From my notes, it says that any convergent sequence is cauchy, so I don't know exactly how to do this. Any suggestions?

    Josh
     
  5. Oct 22, 2006 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you want a sequence that is not Cauchy, and you know that any convergent sequence is Cauchy... then isn't it clear that you should look for a divergent sequence?
     
  6. Oct 22, 2006 #5

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    As Hurkyl hinted, the exercise is aimed at making you think of a sequence that is both bounded and divergent.

    I just got (after 2 years) that when we misleadingly(!) say (for instance in the case of a)) "set u=n²", what we're really doing is arguing that if {[itex]a_n[/itex]} converges to L, then all of its subsequences converge to L, one of which is {[itex]a_{n^2}[/itex]}. And since we know {[itex]a_n[/itex]} to converge to e, the same thing is true of {[itex]a_{n^2}[/itex]}.
     
    Last edited: Oct 22, 2006
  7. Oct 22, 2006 #6

    kreil

    User Avatar
    Gold Member

    How can a sequence be bounded and divergent? If a sequence is bounded then it has a least upper bound s.t. all the elements of the sequence are less than or equal to the l.u.b. If a sequence is divergent, then if you choose any number you can find a number in the sequence that is bigger.

    I think I'm missing something here...
     
  8. Oct 22, 2006 #7

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Nope, don't merge the concepts of divergence and "the limit is [itex]\infty[/itex]". Clearly: a series is divergent iff it does not converge. But that it does not converge does not imediately implies that the sequence in unbounded. There is another possibility...

    <whisper> it's about subsequences! </whisper>
     
    Last edited: Oct 22, 2006
  9. Oct 22, 2006 #8

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you think any bounded sequence converges, then can you give a proof of that claim?

    I'll get you started: Let [itex]\{ s_n \}[/itex] be sequence such that [itex]M \leq s_n \leq N[/itex] for all n.
     
  10. Oct 22, 2006 #9

    kreil

    User Avatar
    Gold Member

    Are we just talking about a sequence whose elements are bounded but when added together are divergent?

    If so, then the harmonic series is bounded but not cauchy, correct?
     
  11. Oct 22, 2006 #10

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Nope; we're talking about sequences, not series.
     
  12. Oct 22, 2006 #11

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The harmonic series is a sequence {[itex]a_n[/itex]} whose general term is the nth partial sum:

    [tex]a_n = \sum_{i=1}^n \frac{1}{i}[/tex]

    And you know that this series diverges because it is unbounded from above. So the divergent series is not what you're looking for.
     
  13. Oct 22, 2006 #12

    kreil

    User Avatar
    Gold Member

    I'm so confused.
     
  14. Oct 22, 2006 #13

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In case of confusion, it is always a good idea to re-read your notes/books (or even better, read the same theory but from another author). Like I hinted, there should be a theorem in there relating the behavior of a sequence to the behavior of its subsequences.
     
  15. Oct 22, 2006 #14

    kreil

    User Avatar
    Gold Member

    I know that if a sequence is bounded then all of its subsequences are bounded by the same bounds. I missed a day of class a few weeks ago which may have been when other notes regarding this were given...
     
  16. Oct 22, 2006 #15

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If a sequence converges to L then all its subsequences converge to L.
     
  17. Oct 22, 2006 #16

    kreil

    User Avatar
    Gold Member

    So, a sequence is convergent iff all of its subsequences converge towards the same limit. And a sequence is bounded iff all of its elements are smaller than or equal to some M>0 for all natural n.

    So a bounded, non-Cauchy sequence is a sequence whose elements are all smaller than or equal to some M>0 for all natural n and whose subsequences don't all converge to the same limit...?

    So would the sequence [itex]a_n={1,-1,1,-1,1,-1.....}[/itex] be a bounded, non-cauchy sequence? It is definitely bounded, and none of its subsequences converge to a limit.
     
    Last edited: Oct 22, 2006
  18. Oct 22, 2006 #17

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    None of its subsequences converge to a limit?! And what about {[itex]a_{2n}[/itex]}={1,1,1,...}?

    That you're on the right track would be an understatement.. you're burning! But you need to find the correct argument.


    P.S. The statement "a sequence is bounded iff all of its elements are smaller than or equal to some M>0 for all natural n." is incorrect. This only says that the sequence is bounded from above, while a bounded sequence is defined as one which is bounded from above and from below. So a standard definition is that {[itex]a_n[/itex]} is bounded if there exist a positive number M such that -M<[itex]a_n[/itex]<M for all naturals n. (Or more compactly that |[itex]a_n[/itex]|<M)
     
  19. Oct 22, 2006 #18

    kreil

    User Avatar
    Gold Member

    not all of its subsequences converge to the same limit, and for that example, M=1. Since it fits this criteria, why is that sequence not bounded and non-Cauchy?
     
  20. Oct 22, 2006 #19

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're asked to find a sequence that is bounded and non-Cauchy.

    In [itex]\mathbb{R}[/itex] a sequence is Cauchy iff it is convergent. So you must find a sequence that is divergent and bounded. Do you see how a_n defined above has these properties?
     
  21. Oct 23, 2006 #20

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Almost giving the answer away: take two sequences that converge to different limits. Make a new sequence by alternating the two convergent sequences.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?