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Convergent sequence

Let \{x_n\}_{n \in \mathbb{N}} be a sequence of positive real numbers such that

(1)   \begin{equation*} x_{n+m} \leq x_n + x_m \quad , \quad  m, n \in \mathbb{N} \end{equation*}

Prove that \left\{\dfrac{x_n}{n} \right\}_{n \in \mathbb{N}} converges.


Fix m and let n \geq m. Then, there exist k, r such that n=km+r where 0\leq r <m. Thus,

    \[\frac {x_n}{n} = \frac {x_ {km+r}}{n}\leq \frac {kx_ {m}}{n} + \frac {x_{r}}{n}\]

Letting n \rightarrow +\infty it follows that

    \[\limsup \frac {x_n}{n} \leq \frac {x_ {m}}{m} + 0\]

Since this holds forall m it follows that \displaystyle \limsup \frac {x_n}{n} \leq \liminf \frac {x_ {m}}{m} and the result follows.

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On a nested sin sequence

Consider the sequence x_n defined recursively as

    \[x_1=1 \quad, \quad x_{n+1}=\sin x_n\]

Prove that \lim \limits_{n \rightarrow +\infty} \sqrt{n} x_n = \sqrt{3}.


Lemma: If a_n is a sequence for which \displaystyle \lim_{n\to+\infty}(a_{n+1}-a_n)=a then

    \[\lim_{n\to + \infty}\frac{a_n}n=a.\]

Proof: In Stolz theorem we set x_{n}=a_{n+1} and y_n=n.

It is easy to see that x_n is is monotonically decreasing to zero. Moreover, an application of L’Hospital’s rule gives

    \[\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac{1}{3}\]



Now, due to the lemma we have \lim\limits_{n\to+\infty} na_n^2 = 3 and the result follows.

Remark : The asymptotic now follows to be \displaystyle x_n \sim \sqrt{\frac{3}{n}}.

Problem: Find what inequality should \beta satisfy such that the series

    \[\mathcal{S}=\sum_{n=1}^{\infty} x_n^\beta\]


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On the supremum and infimum of a sine sequence

Let \{x_n\}_{n \in \mathbb{N}} be a sequence defined as

    \[x_n = \sin 1 + \sin 3 + \sin 5 + \cdots + \sin (2n -1)\]

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


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.

    \begin{align*} \sum_{k=1}^n\sin (2k-1) &= {\rm Im}\left[\sum_{k=1}^ne^{i(2k-1)}\right]\\ &= {\rm Im}\left[\frac{e^{i}(1-e^{2ni})}{1-e^{2i}}\right]\\ &= {\rm Im}\left[\frac{e^{i}}{1-e^{2i}}(1-e^{2ni})\right]\\ &= {\rm Im}\left[\frac{e^{i}}{1-e^{2i}}(1-\cos 2n+i\sin 2n)\right]\\ &= {\rm Im}\left[\frac{i}{2\sin 1}(1-\cos 2n+i\sin 2n)\right]\\ &=\frac{1-\cos 2n}{2\sin 1} \end{align*}

Thus \displaystyle x_n= \frac{1-\cos 2n}{2\sin 1} and we have to find the supremum and infimum of \cos 2n. Since the values n \mod 2\pi are dense on the unit circle , the same shall hold for 2 n \mod 2\pi implying that \inf \cos 2n =-1 and \sup \cos 2n = 1. Thus,

\displaystyle \inf\{x_n\}=\frac{1}{2\sin 1}(1-1)=0 \quad , \quad \sup\{x_n\}=\frac{1}{2\sin 1}(1+1)=\frac{1}{\sin 1}

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Unbounded sequence

A sequence of real number \{x_n\}_{n \in \mathbb{N}} satisfies the condition

    \[|x_n-x_m| > \frac{1}{n} \quad \text{whenever} \quad n<m\]

Prove that x_n is not bounded.



    \[I_n= \left( x_n - \frac{1}{2n} , x_n + \frac{1}{2n} \right)\]

The inequality \left| x_n - x_m \right| > \frac{1}{n} for n<m implies that I_n \cap I_m = \varnothing for m \neq n. Further, let I = \bigcup \limits_{n=1}^{\infty} I_n. The sum of lengths of all the intervals I_n for n=1, 2, \dots is clearly infinite and hence I is unbounded and so is the sequence x_n.

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