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Basic sequence help. (Convergence)

  1. Oct 24, 2011 #1
    Im struggling with the concept of this basic sequence question.

    Let x(n) be a sequence such that lim(n->00) (nx(n)) = 0

    i.e. it converges to zero...

    How could i show that there is an N s.t. for all n≥N : -1 < nx(n) < 1

    Any tips would be great.. I don't want an answer.. I want to be guided through it please.

    Regards as ever.
  2. jcsd
  3. Oct 24, 2011 #2
    What is the definition (using [itex]\epsilon[/itex] and N) of convergence for a sequence? Once you've written that, it should be straightforward to answer your question.
  4. Oct 25, 2011 #3
    the convergence of a seq:

    A sequence {r^n}00n=0 converges if -1< r <1

    Attempt at answer:

    Same as above but replace 'r^n' with 'nx(n)

    Thanks so far
  5. Oct 25, 2011 #4
    ^ Wut?? That's the result of convergence of a geometric series, it's very different.

    A sequence of real numbers [itex](x_n)[/itex] converges when there exists a number [itex]L \in \mathbb{R}[/itex] such that, for any [itex]\epsilon > 0[/itex], I can find a number [itex]N \in \mathbb{N}[/itex] so that [itex]|x_n - L| < \epsilon[/itex] whenever [itex]n > N[/itex]. This is very wordy but I'm sure this was introduced in your math class. The number L is the limit of the sequence [itex](x_n)[/itex].

    What do you mean by "nx(n)"? Do you mean [itex]nx_n[/itex]?? If so, to show your desired result, use the definition from above. We know that for any number [itex]\epsilon > 0[/itex], I can find [itex]N \in \mathbb{N}[/itex] so that [itex]|nx_n - L| < \epsilon[/itex] whenever n > N. Fill in the details now: we know L = 0. Also, the previous statements work for any chosen [itex]\epsilon[/itex] -- how can you choose [itex]\epsilon[/itex] to get the result?
  6. Oct 26, 2011 #5

    |x(n) - L| < ε

    therefore: |nx(n) - 0| < ε

    therefore: nx(n) <

    so, -ε < nx(n) < ε

    so, take epsilon to be ε=1 and we have: -1 < nx(n) < 1

    How does this look guys?

    Thanks so far
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