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Proving sequences

  1. Dec 6, 2011 #1
    if lim(n→∞)an=0 then prove lim(n→∞)[itex]1/an[/itex]=0

    how do i do this, i know how to proove it geometrically, but how do you write the proof using ε
    and [itex]\delta[/itex]

    Give a counter example to show that the converse is false.
     
    Last edited: Dec 6, 2011
  2. jcsd
  3. Dec 6, 2011 #2

    Erland

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    There must be a typo here, for if an -> 0 then obviously 1/an -> infinity (if an>0, otherwise it may oscillate between +infinity and -infinity).
     
  4. Dec 6, 2011 #3
    if lim(n→∞) an = ∞ then prove lim(n→∞) 1/an = 0 is what it shoud say say sorry. where an is a sequence
     
  5. Dec 6, 2011 #4
    What have you tried? We can't help until we see your attempt. The only hint I will give is to write out the definitions. What does [itex] \lim_{n \to \infty} a_n = \infty [/itex] mean? How about [itex] \lim_{n \to \infty} 1/a_n = 0 [/itex]?
     
  6. Dec 6, 2011 #5
    Why don't you first prove that the limit of 1/n as n goes to infinity is zero, and then consider the limit of 1/a_n as a composition?
     
  7. Dec 7, 2011 #6
    i can prove it graphically but i dont know the deffinitions and this is not for assingnment, once someone shows me how its done then i can start my assignments, i need the basics first
     
  8. Dec 7, 2011 #7
    You can find the necessary definitions here: http://en.wikipedia.org/wiki/Limit_of_a_sequence
     
  9. Dec 7, 2011 #8
    thanks. can you please show me how to aproach it
     
  10. Dec 7, 2011 #9
    Here's the definition for [itex] \lim_{n \to \infty} a_n = \infty [/itex]. For any [itex]M > 0[/itex] there is positive integer N such that for any [itex] n \geq N [/itex], [itex] a_n > M [/itex].

    Now you write out the definition for [itex]\lim_{n \to \infty} 1/a_n = 0 [/itex] and try to see why [itex] \lim_{n \to \infty} a_n = \infty [/itex] implies [itex]\lim_{n \to \infty} 1/a_n = 0 [/itex] from these definitions.
     
  11. Dec 7, 2011 #10
    For any ε positive you should find an N s.t for all n>N |1/an - 0| < ε. You must show that it would suffice to do this for all ε = 1/m where m is an integer (exercise). Then since your sequence goes to infinity, you can find an N s.t for all n>N an > m hence 1/an < 1/m and then you are done (fill in the gaps if you want to understand the definition of limit).
     
  12. Dec 20, 2011 #11
    thanks for this, can you please give a counter example to show the converse is false?
     
  13. Dec 21, 2011 #12

    I think the converse is also true,here is a proof:

    Let M>0,then (1/M)>0

    Since [tex]lim_{n\to\infty}\frac{1}{a_{n}}=0[/tex] for all ε>0 there exists a natural No k such that:

    for all ,n [tex]n\geq k\Longrightarrow \frac{1}{a_{n}}<\epsilon[/tex].

    Put [tex]\epsilon = \frac{1}{M}[/tex] and we have that :

    for all ,n [tex] n\geq k\Longrightarrow \frac{1}{a_{n}}<\frac{1}{M}\Longleftrightarrow a_{n}>M[/tex]

    Hence [tex]lim_{n\to\infty} a_{n} = \infty[/tex]
     
  14. Dec 21, 2011 #13

    micromass

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    This step is false:

    [tex]\frac{1}{-2}<\frac{1}{2}[/tex]

    but not

    [tex]-2>2[/tex]
     
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