On a nested sin sequence

Consider the sequence defined recursively as

Prove that .

Solution

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

Therefore

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

converges.

On the supremum and infimum of a sine sequence

Let be a sequence defined as

Find the supremum as well as the infimum of the sequence .

Solution

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,

Unbounded sequence

A sequence of real number satisfies the condition

Prove that is not bounded.

Solution

Let

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 .