Proof for "A sequence tends to infinity if, as n increases, $a_n$ increases without bound. Formulate a precise definition." This is the full question:

My attempt at solving this is:
Precise definition: To say that $\{n\}^\infty_{n=1}$ tends to infinity is to say that $\frac{1}{n}$ gets closer to zero as n gets closer to $\infty$.
Proof: Let $\epsilon \gt 0$. We pick an $n\gt\frac{1}{\epsilon}$, so $\frac{1}{n}\lt\epsilon$. For all $m \ge n, \frac{1}{m} \le \frac{1}{n} \lt \epsilon$ which proves $\frac{1}{n}$ gets closer to zero.
The second sequence follows the precise definition, but I have confusion about the proof.
My attempt: Let $\epsilon \gt 0$ and $n \gt \frac{1}{\epsilon}$, so $\frac{1}{n} \lt \epsilon$. For all $m \ge n, \frac{1}{2^m} \lt \frac {1}{m} \le \frac {1}{n} \lt \epsilon$, which proves $\frac{1}{2^n}$ tends to zero.
My question is, is the way I made the precise definition correct, if not, how to make it? Is there any source I should read?
Secondly, is my second proof correct? If not, what would be the correct proof?
Thank you for reading.
 A: The problem with your first precise definition is that you're saying $n$ tends to infinity if $n$ gets closer to infinity. This isn't particularly useful.
The usual way we define a limit as being infinite is by altering the $\epsilon$ definition of a limit. We say a sequence $(a_n)$ tends to infinity if for all $M$ there exists an $N$ such that for all $n\geq N$ we have that $a_n\geq M$. Clearly this holds for the sequence given by $a_n=n$ as we can take $N=M$.
In the case of the sequence $a_n=2^n$ we could again take $N=M$ as $2^n\geq n$ for all $n\geq0$.
A: A good attempt at a definition, but it unfortunately oversees an important requirement of your sequence. You say that your sequence is increasing if the reciprocals of the terms of the sequence get closer and closer to $0$. Now, consider the following sequence:
$$9,9.9,9.99,9.999,\ldots$$
The reciprocals of the terms are:
$$\frac{1}{9},\frac{10}{99},\frac{100}{999},\frac{1000}{9999},\ldots$$
You can see that all of these reciprocals get smaller  (and hence closer to $0$). But, they have the values:
$$0.111111\ldots , 0.101010\ldots , 0.100100 \ldots , \ldots$$
None of these reciprocals are going to be less than $0.1$. Basically, no term in your original sequence is greater than $10$. So, your sequence does increase, but with bound.
To ensure your sequence increases without bound, you must not only ensure that your sequence increases, but your sequence must increase such that it can get arbitrarily large. Can you continue from here?
A: The informal definition is wrong because

*

*there are non-increasing sequences that do tend to infinity, and


*there are sequences that do not tend to infinity though they are unbounded above.
A correct definition is that taking a bound as large as you want, you can find a tail of the sequence that stays above it (i.e. some $a_n$ and all following elements).
$$\forall m:\exists n:\forall k\ge n:a_k\ge m.$$
Sequence $n$:
With $n=m$,
$$\forall m:\forall k\ge m:a_k=k\ge m.$$
Sequence $2^n$:
Also with $n=m$,
$$\forall m:\forall k\ge m:a_k=2^k\ge m.$$
