Does Sequence (n,1/n) Converge or Diverge?

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Discussion Overview

The discussion centers on the convergence or divergence of the sequence (n, 1/n) as n approaches infinity. Participants explore the behavior of the sequence in the context of convergence in R², considering the limits of its components.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant suggests that the sequence (n, 1/n) converges, noting that as n approaches infinity, 1/n approaches 0.
  • Another participant questions the convergence, stating that for a sequence to converge in R², both components must converge separately, leading to the conclusion that (n, 1/n) diverges since n diverges.
  • A later reply clarifies that the sequence (1, 1/n) converges, while (n, 1/n) diverges, indicating a distinction between the two sequences.
  • Some participants express gratitude for the clarification and acknowledge the confusion regarding the notation used.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the convergence of (n, 1/n), with some arguing it diverges and others suggesting it converges based on different interpretations of the sequence.

Contextual Notes

There is a potential misunderstanding regarding the notation of the sequences discussed, which may affect the conclusions drawn by participants. The distinction between (n, 1/n) and (1, 1/n) is noted but not fully resolved.

zendani
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if we have a sequence (n,1/n) , n E N , the sequence converges?

lim n = infinite
lim 1/n = 0

(1,1),(2,1/2),(3,1/3)...(n,1/n)

it is convergent and divergent?!
 
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if anybody knows about such a sequence, book or reference, please write here

because i want to learn it

Thank you
 
zendani said:
if anybody knows about such a sequence, book or reference, please write here

because i want to learn it

Thank you

In order to converge in R^2, the x-y plane, a sequence of points has to converge in each variable separately. So the sequence (1, 1/n) does not converge.
 
For a sequence of the form (xn,yn) to converge, we require that both xn and yn converges. Here, xn=n, yn=1/n. While yn converges to 0, xn diverges so we say that (n,1/n) diverges.
 
thank you Stevel27 and quasar987, i got it

stevel, i have (n,1/n) no (1,1/n)

so (n, 1/n) diverges and (1,1/n) converges...
 
correct! :)
 
zendani said:
thank you Stevel27 and quasar987, i got it

stevel, i have (n,1/n) no (1,1/n)

so (n, 1/n) diverges and (1,1/n) converges...

Yes, you're right about that. Typo on my part, but of course (1, 1/n) does converge.
 

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