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Let be a sequence of positive real numbers such that
Prove that converges.
Fix and let . Then, there exist such that where . Thus,
Letting it follows that
Since this holds forall it follows that and the result follows.
Consider the sequence defined recursively as
Prove that .
Lemma: If is a sequence for which then
Proof: In Stolz theorem we set and .
It is easy to see that is is monotonically decreasing to zero. Moreover, an application of L’Hospital’s rule gives
Now, due to the lemma we have and the result follows.
Remark : The asymptotic now follows to be .
Problem: Find what inequality should satisfy such that the series
Let be a sequence defined as
Find the supremum as well as the infimum of the sequence .
Background: This problem was on the shortlist of the 2014 Olimpiada Nationala de Matematica de Romania and was suggested by Leo Giugiuc.
We begin by the very well known manipulation.
Thus and we have to find the supremum and infimum of . Since the values are dense on the unit circle , the same shall hold for implying that and . Thus,
A sequence of real number satisfies the condition
Prove that is not bounded.
The inequality for implies that for . Further, let . The sum of lengths of all the intervals for is clearly infinite and hence is unbounded and so is the sequence .