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Finding an oscillating sequence that diverges and whose limit is zero.

  • Thread starter 4zimuth
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  • #1
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Homework Statement



Hi, I need to find an oscillating sequence whose limit of the differences as n approaches infinity is zero but the sequence itself is diverging.

Homework Equations



None.

The Attempt at a Solution



My initial guess was:

[tex]\frac{sin(ln(n))}{ln(n)}[/tex]
 

Answers and Replies

  • #2
HallsofIvy
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What do you mean by an "oscillating" sequence?
 
  • #3
Hi!

If I understand correctly, then
[tex]sin\left(\sqrt{n}\right)[/tex]
is what you are looking for. The proof is based on the idea sin(a)-sin(b) < a-b and sqrt(n+1)-sqrt(n) converges to 0.
I originally bumped this post, because I need help with a similar problem, where not
an+1-an converges to 0, but an+1/an converges to 1.
Does anybody know a good example for the latter? So the problem again:
We need a sequence, that:
1) oscillates (oscillation is when it divergates, but neither to infinity nor negative infinity, eg. -1 +1 -1 +1 ... ; +1 -2 +3 -4 +5 -6 ... ;sin(n) ; etc.)
2) an+1/an converges to 1
 

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