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Homework Help: Analysis Question: Proving Algebraic operations for infinite limits

  1. Mar 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Given [tex]\lim_{n\rightarrow \infty}a_{n}}= 0[/tex]
    [tex]b_{n}[/tex] is bounded below.

    Prove: [tex]\lim_{n\rightarrow \infty}(a_{n}+b_{n})}= \infty[/tex]

    2. Relevant equations

    3. The attempt at a solution

    According to my text: [tex]{b_{n}}[/tex] is bounded below if and only if there is a real number [tex]\ni[/tex] B [tex]\leq[/tex] [tex]b_{n}\forall_{n}[/tex]

    So, here's my attempt:

    Putting the givens together I get:

    B [tex]\leq[/tex] [tex]b_{n}[/tex] [tex]\leq[/tex] 0

    At this point forward I'm not sure where to go with this. Any kind of help is appreciated.
  2. jcsd
  3. Mar 3, 2010 #2

    Gib Z

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    Homework Helper

    You can't prove it as it is written, as its not true. For example, the sequences [itex]a_n = 0[/itex] and [itex]b_n = 1[/itex] satisfy the given conditions, yet the limit of their sum is 1, not divergent to infinity.

    Are you sure thats exactly how the question is given? Perhaps they were sloppy and by their statement "Bounded from below" they also implied "Unbounded from above".

    In the likelyhood the question is the case, Use that fact. What does it mean for a sequence to be unbounded from above?
  4. Mar 3, 2010 #3
    That's exactly how the given was quoted in my textbook. That's why I was confused as well.

    If a sequence is unbounded from above then it goes to infinity.
  5. Mar 3, 2010 #4

    Gib Z

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    Homework Helper

    Well yes but in algebraic term. You wrote "According to my text: [tex]{b_{n}}[/tex] is bounded below if and only if there is a real number [tex]\ni[/tex] B [tex]\leq[/tex] [tex]b_{n}\forall_{n}[/tex]".

    Write the corresponding statement for if a sequence is unbounded from above. There will be information in there that you can use.

    Next thing to do would be a proof by contradicition, assume the [itex]\lim_{n\to \infty} a_n +b_n[/itex] does exist and is equal to some number L. By the epsilon delta definition of a limit would does it mean for the limit to be equal to L? Can you contradict that using any information you have
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