Prove that a positive, null sequence has a maximum

Question

I'm looking for some insight on how to formalize my ideas for this proof.

So because the sequence is null, we know that the sequence taper off to $0$ for some $n$ values beyond $N$. We can therefore use $N$ to divide the sequence into a finite, non-empty part (the part before $N$) and an infinite, bounded part (the part after $N$).

There is a theorem stating that every finite, non-empty set has a maximum, so we know that the part of the sequence before $N$ has a maximum. Since we know that the part after $N$ tapers off to $0$, we know that this maximum value applies to this part of the sequence too. Thus, every positive, null sequence has a maximum.

Answer 1

All your arguments can be formalised. Let's call the sequence $(x_n)_n$.

First, you split the sequence into a finite part and a tail. We can do this a bit more explicitly: let $\epsilon = 1$, then we can find a natural $N$ such that for every natural $n>N$, we have $\lvert x_n - 0\rvert < \epsilon$, which means every element in the tail of the sequence is smaller than $\epsilon$.

Now, we know that $M := \max\{x_1, x_2, \dots, x_N\}$ exists, because the maximum is taken over a finite amount of (finite) numbers. This is probably close to the theorem you stated.

We can finish the proof by noting that every element of the sequence is smaller than $\max\{\epsilon, M\}$.

Note that you stated that since the sequence tapers off to 0, every element in the tail is also smaller than $M$. This is only the case if we manage to choose the right $\epsilon$. Can you see why?

Answer 2

You have almost finished the proof: Choose $n_0$ such that $a_n <a_1$ for $n >n_0$. Then the maximum of the numbers $a_1,a_2,...,a_{n_0}$ is also the maximum of the entire sequence.

I am assuming that positive means strictly positive. If zeros are allowed then the maximum of the nonzero terms of the sequence (if any) is attained and this gives the maximum of the sequence.

Answer 3

Pretty much you got the idea.

If all terms are $0$, done.

Otherwise, take some $a_{k}= m >0$. Since the sequence converges to $0$ there exists $N$ so that $a_n < m$ for $n> N$. Notice that we must have $N\ge k$. Now take $n_0\in \{1, \ldots N\}$ the largest of the $a_i$, $1\le i\le N$. Certainly $a_{n_0}\ge a_k = m$. Now all the terms of index $>N$ are $< m$ so $< a_{n_0}$. Therefore, $a_{n_0}= \max_n \{a_n\}$

Answer 4

An exercise using a few basic ideas.

$(a_n)$ positive, converges to $0$.

Since $(a_n)$ is convergent, it is bounded.

Let $A:=\{a_n\}$ be the underlying set, and

$L:=\sup \{a_n\}$;

1) If there is a $n_0$ s.t. $a_{n_0} =L$, we are done.

2) Assume for all $n \in \mathbb{N}: L> a_n$.

$L=\sup_{n} \{a_n\}$, there is a sequence of distinct $b_k\in A$ converging to $L$, since:

Since $L-1/k$, $k\in \mathbb{N}$ is not an

upper bound, there is a $b_k$ s.t. $L> b_k >1/k$.

$b_k$ converges to $L$.

It follows there are Infinitely many distinct $b_k$ in a sufficiently small neighbourhood of $L>0$, a contradiction to $\lim_{n \rightarrow \infty}a_n=0$ (Why?)

Hence only option 1).