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

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An oscillating sequence that diverges while having its limit of differences approach zero can be represented by sin(√n). The discussion emphasizes the definition of oscillation, which involves divergence without approaching infinity. Additionally, a related problem is introduced, seeking a sequence where the ratio of consecutive terms converges to 1. The participants explore various examples and clarify the conditions for oscillation. The conversation highlights the complexity of finding such sequences in mathematical analysis.
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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:

\frac{sin(ln(n))}{ln(n)}
 
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What do you mean by an "oscillating" sequence?
 
Hi!

If I understand correctly, then
sin\left(\sqrt{n}\right)
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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