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

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.

• so {a_n} is finite, which is a contradiction Why would that be a contradiction? – dxiv Sep 26 '18 at 0:25
• $\{0,0,\cdots \}$ is such a sequence. – Kabo Murphy Sep 26 '18 at 0:26
• 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. – Alana Sep 26 '18 at 0:26
• @Alana Take any strictly increasing convergent sequence $a_n\to A$, pick a $B \ge A$, then replace $a_1$ with $B$. – dxiv Sep 26 '18 at 0:58
• A note on why you must pick a sequence which converges $a_n\to A$ is due to the monotone convergence theorem – JMoravitz Sep 26 '18 at 1:00

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 $$\text{max}\{a_n\} \geq \text{max}\{a_n\}_{n \leq 1} = a_1 > a_0$$.