Basic sequence help. (Convergence)

  • Thread starter tamintl
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  • #1
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Main Question or Discussion Point

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.
Tam
 

Answers and Replies

  • #2
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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.
 
  • #3
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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
 
  • #4
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^ 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?
 
  • #5
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Right:

|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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