If $a_n$ is a strictly monotonic-increasing sequence of integers then $\lim_{n\to\infty}a_n=\infty$

Question

As shown in the title, I'm asked to prove that such sequence diverges to infinity.
My attempt:
Given $a_n$ is monotonic increasing of sequence of integers implies it's not bounded above and the subtraction of two consecutive elements is at least $1$. Thus we obtain:
$a_2-a_1\ge 1 \Longleftrightarrow a_2\ge a_1+1$
$a_3-a_2\ge 1 \Longleftrightarrow a_3\ge a_2+1$
This yields $a_3\ge a_2+1\ge a_1 +2$.
If we continue inductively we obtain $a_{n+1}\ge a_1+n$.
This inequality yields the following limit inequality $\lim_{n\to\infty}a_{n+1}\ge \lim_{n\to\infty}a_1+n$.
Calculate the RHS limit: $\lim_{n\to\infty}a_1+\lim_{n\to\infty}n=a_1+\infty=\infty$.
Now using limit comparison, $\lim_{n\to\infty}a_{n+1}=\infty$.
I'd like to hear some feedbacks because I'm not sure about the inductive construction and the bounded above claim.

Answer 1

You want to show that

$$\forall M> 0\, \exists N\in\mathbb{N}, n > N: |a_n| > M $$

and I assume that you already proved by induction that $$\forall n\in \mathbb{N}: a_{n} \ge (n-1)+a_1$$

Let's choose an arbitrary large $M>0$. By the Archimedian property there exists a natural number $$N>max(M+1-a_1,1)\ge M+1-a_1,$$ such that for any $n>N$ it holds

$$a_n\ge n-1+a_1>N-1+a_1>M$$

Answer 2

The inductive formation and conclusion of the claim $a_{n+1} \ge a_1+ n$ for all terms of the sequence is just fine.

(Base step: For $n=1$: $a_{2} > a_1$ so $a_2-a_1 > 0$ but $a_2-a_1\in \mathbb Z$ and there are no integers between $0$ and $1$ so $a_2 -a_1\ge 1$. Induction step: If $a_{n+1}\ge a_n + n$ then $a_{n+2}> a_{n+1}$ and by the same argument as the base case $a_{n+2} - a_{n+1} \ge 1$. So simple manipulation yields $a_{n+2} \ge a_{n+1} + 1 \ge (a_1 + n) + 1 =a_1 + (n+1)$.)

(Although maybe the claim is more flexible if stated as $a_n \ge (a_1-1) + n$)

Then your intuition is basically correct: If $a_{n+1} \ge a_1 + n$ then as $n\to \infty$ we must have $a_{n+1} \ge a_1 + n$ must also tend to infinity. But at this point you should state it formally and properly.

You make a few casual hand-wavy actions that are not really acceptable.

One: You treat $\infty$ as though it is a number. It isn't really. It is a symbol we can just to have to mean certain properties. Two: you assume $\lim_{n\to \infty} a_{n+1}$ exists (or is expressible) without really verifying it. Three: You claim $\lim (a_n + n) = \lim a_n +\lim n$ without verifying such limits exits or are expressible. Four: you use the expression $b + \infty=\infty$ as though that is a meaningful expression.

I'd suggest using the definition of $\lim_{n\to \infty} b_n = \infty$ direct: If for any $M$ there exists an $N\in \mathbb N$ so that for all $n\ge N$ we have $a_n >M$ we say $\lim_{n\to \infty}a_n = \infty$.

This makes for a very straightforward proof. Our only issue is if we choose an arbitrary $M$ how do we choose an $N$ so that $n \ge N$ implies $a_n \ge M$.

Well, we do some calculations on scratch paper. We have $a_n \ge a_1 + (n-1)$ always. And if $n \ge N$ we have $a_n \ge a_1 + (n-1) \ge a_1 + (N-1)$. We WANT that to imply $a_n \ge M$ and that will be true if we have $a_n \ge a_1 + (N-1) > M$. So what value of $N$ will make that true?

Well $a_1 + (N-1) > M\iff N > M+1 -a_1$.

So that's our proof. Let's write it out.

Let $M$ be any arbitrary value (presumably as large as we'd like). Let $N$ be any natural number so that $N > M+1 - a_1$. The for any $n\ge N$ we have by our initial claim that $a_n\ge a_1 + (n-1)$. And using $n \ge N > M+1 - a_1$ we have:

$a_n\ge a_1+(n-1)\ge a_1 + (N-1) > a_1 + ((M+1-a_1)-1) = M$.

Thus the criteria for $\lim_{n\to \infty} a_n = \infty$ are fulfilled.

POST-SCRIPT:

There are, depending on your text, some useful properties we can use. Two key might be:

1. If $a_n$ and $b_n$ are such that $a_n \ge b_n$ and $\lim_{n\to \infty}b_n=\infty$ then $\lim_{n\to \infty} =a_n = \infty$.

Pf: $\lim_{n\to \infty}b_n=\infty$ so for any $M$ there is an $N$ so that if $n\ge N$ then $b_n > M$. But then if $n\ge N$ then $a_n \ge b_n > M$. So $\lim_{n\to \infty} a_n = \infty$.

2. If $\lim_{n\to \infty} a_n = L$ and $\lim_{n\to \infty}b_n = \infty$ then $\lim_{n_\to\infty} (a_n + b_n)=\infty$

Pf: Let $M$ be any value.

$\lim_{n\to \infty}a_n =L$ so for any $\epsilon > 0$ there is an $N$ so that if $n>N$ we have $|a_n - L| < \epsilon$. Let $N_1$ be the value so that $n \ge N_1$ implies $|a_n -L| < 1$.

$\lim_{n\to \infty}b_n = \infty$. Let $M_1 = M-L + 1$. There exist an $N_2$ so that $n \ge N_2$ implies $b_n > M_1$.

Let $N = \max(N_1,N_2)$. If $n \ge N$ then $n\ge N_1$ and $|a_n - L|< 1$ so so $-1< a_n - L < 1$ so $a_n > L -1$. And $n \ge N_2$ so $b_n > M-L + 1$. So $a_n + b_n > (L-1)+M-L + 1= M$.

So $\lim_{n\to \infty} (a_n + b_n)=\infty$.

...

Then as $a_n \ge a_1 + (n-1)$ and as $\lim_{n\to \infty} a_1 = a_1$. (Pf: For any $\epsilon > 0$ if $n> 0$ then $|a_1-a_1|=0< \epsilon$) and as $\lim_{n\to \infty} (n-1)=\infty$ (Pf: For any $M$ if $n\ge M+2$ then $n-1\ge M+1 > M$), we have $\lim_{n\to \infty} a_1 + (n-1) = \infty$ and so $\lim_{n\to\infty}a_n = \infty$.