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Write down an example of a sequence ${a_n}$ such that:

${a_n}$ is divergent

$a_n < 5$ for all $n$, and

$a_n < a_{n+1}$ for all $n$,

or explain why no such divergent sequence exists.


What I did thus far:

We know that the sequence is increasing since we are given that $a_n < a_{n+1}$ for all $n$, thus we can also say that it is monotonic.

We also know that the sequence is bounded from above since we are given that $a_n < 5$ for all $n$.

By the Monotonic Sequence Theorem, we know that every bounded and monotonic sequence is convergent.

Thus, no such divergent sequence exists.

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    Looks OK to me. – marty cohen Nov 15 '16 at 01:07
  • I dont know the background of the problem but the answer is incomplete. Observe that the sequence defined in $\Bbb Q$ by $$x_{n+1}=\frac{4+3x_n}{3+2x_n},\quad\text{with }x_0=1$$ converges monotonically from below to $\sqrt 2$ what doesnt exists in $\Bbb Q$, so it is not a convergent sequence in $\Bbb Q$, but it converges in $\Bbb R$, due to the fact that $\Bbb R$ is complete but not $\Bbb Q$. Then I assume that the context of this exercise is $\Bbb R$. – Masacroso Nov 15 '16 at 03:39

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