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

4zimuth
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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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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