Exercise 2.6.6

Question

Let's call a sequence $\left(a_{n}\right)$ quasi-increasing if for all $\epsilon>0$ there exists an $N$ such that whenever $n>m \geq N$ it follows that $a_{n}>a_{m}-\epsilon$
  \begin{enumerate}[label=(\alph*)] 
  \item Give an example of a sequence that is quasi-increasing but not monotone or eventually monotone.
  \item Give an example of a quasi-increasing sequence that is divergent and not monotone or eventually monotone.
  \item Is there an analogue of the Monotone Convergence Theorem for quasiincreasing sequences? Give an example of a bounded, quasi-increasing sequence that doesn't converge, or prove that no such sequence exists.
   \end{enumerate}

Ulisse Mini · Accepted Answer

Think of ``quasi-increasing'' as ``eventually the $n$'th term will be almost smaller then all terms after it'' \begin{enumerate}[label=(\alph*)] \item $a_n = (-1)^n/n$ is quasi-increasing since $(-1)^m/m - (-1)^n/n \le 1/m + 1/n < 2/N < \epsilon$ after picking some $N > 2/\epsilon$. \item $a_n = (2,2-1/2,3,3-1/3,\dots)$ is quasi-increasing. Let $\epsilon > 0$ and set $N > 1/\epsilon$, for $n > m \ge N$ consider two cases We have $a_n > a_m$ as long as $a_n e m-1/m$. If $a_n = m-1/m$ then $a_n > a_m-\epsilon$ since $m-1/m > m-\epsilon$ as $1/m < \epsilon$. \item Suppose $(a_n)$ is quasi-increasing and bounded and let $\epsilon > 0$. Let $N_1$ be large enough that $n > m \ge N_1$ implies $a_n > a_m - \epsilon$. Since $(a_n)$ is bounded we can set $s = \sup a_n$ applying Lemma 1.3.8 tells us there exists an $N > N_1$ such that $a_N > s-\epsilon$. Now for all $n > N$ we have $a_n > a_N-\epsilon$, and since $a_N > s-\epsilon$ we have $a_n > s-2\epsilon$. This completes the proof as $s \ge a_n > s-2\epsilon$ implies $|a_n - s| < 2\epsilon$ \emph{for all} $n \ge N$, thus $s = \lim a_n$. \end{enumerate}

Exercise 2.6.6

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Exercise 2.6.6

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