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Edit 1: Changed from monotonically increasing sequence to a strictly increasing sequence.

Edit 2: Let $\{a_n\}$ denote a sequence such that for all $n\geq 2$, $a_n < a_{n+1}$ and $a_n \leq a_1$.

Is it possible to have a sequence that is strictly increasing but whose first term is the maximum?

I think not.

Let $\{a_n\}$ be a sequence such that for all $n\geq 2$ we have $$a_n < a_{n+1}$$ Suppose that $a_n \leq a_1$ for all $n\geq 2$. Then $\{a_n\}$ is an infinite increasing sequence that has a maximum, so $\{a_n\}$ is finite, which is a contradiction.

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  • $\begingroup$ so {a_n} is finite, which is a contradiction Why would that be a contradiction? $\endgroup$ – dxiv Sep 26 '18 at 0:25
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    $\begingroup$ $\{0,0,\cdots \}$ is such a sequence. $\endgroup$ – Kabo Murphy Sep 26 '18 at 0:26
  • $\begingroup$ Because $\{a_n\}$ is a sequence, which is a function whose domain is the set of natural numbers. Thus there are as many terms in the sequence as there are natural numbers, so $\{a_n\}$ is an infinite sequence. But I've shown that $\{a_n\}$ is finite, which is a contradiction. $\endgroup$ – Alana Sep 26 '18 at 0:26
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    $\begingroup$ @Alana Take any strictly increasing convergent sequence $a_n\to A$, pick a $B \ge A$, then replace $a_1$ with $B$. $\endgroup$ – dxiv Sep 26 '18 at 0:58
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    $\begingroup$ A note on why you must pick a sequence which converges $a_n\to A$ is due to the monotone convergence theorem $\endgroup$ – JMoravitz Sep 26 '18 at 1:00
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As Kavi Rama Murphy pointed out in his comment, $a_n = 0$ satisfies the criterion you described. However,it is true that there exists no strictly increasing sequence with more than a single element for which $a_0 = \text{max}\{a_n\}$ is the maximum. In particular, because $a$ is strictly increasing, $a_0 < a_1$, which implies that $\text{max}\{a_n\}_{n \leq 1} = a_1 > a_0$. Further, since $\{a_n\}_{n \leq 1} \subset \{a_n\}$, we have $\text{max}\{a_n\} \geq \text{max}\{a_n\}_{n \leq 1} = a_1 > a_0$.

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