Answer
False
Work Step by Step
Step 1:
We have to prove or disprove that whenever a sequence diverges, its terms must either approach positive or negative infinity.
Step 2
This statement is false and we can just give a counterexample that supports this claim. Let's observe the sequence with general term $a_n = (-1)^n$. First we will write several terms of this sequence and see if we can conclude something about the convergence.
$a_1=(-1)^1=-1, a_2=(-1)^2=1, a_3=(-1)^3=-1, a_4=(-1)^4=1, \ldots$
As we can see, all the terms with odd indices will be equal to $-1$ and all the terms with even indices will be equal to $1$.
What can we conclude about the convergence?
Step 3
There is a theorem that states that a sequence converges if and only if the two subsequences, one with even and the other with odd index terms, converges and their limits are the same.
Step 4
Here we have that the subsequence with even index terms is ${a_{2k}}_{k\in\mathbb{N}}$ and its limit is
$\displaystyle\lim_{k\to\infty}a_{2k}=\lim_{k\to\infty}1=1.$
On the other hand, the subsequence with odd index terms ${a_{2k-1}}_{k\in\mathbb{N}}$ converges to a different value:
$\displaystyle\lim_{k\to\infty}a_{2k-1}=\lim_{k\to\infty}-1=-1.$
Therefore, we have that the original sequence does not converge, so it diverges, but the terms do not approach neither $+\infty$ nor $-\infty$.
Step 5
We will now state more counterexamples that prove this is a false statement.
Observing the sequence $a_n=(-1)^n n$, we have that one subsequence, with even index terms, will approach $+\infty$ and the other one, with odd index terms, will approach $-\infty$. Since those two are not the same, we have that the limit of ${a_n}$ does not exist as $n\to\infty$.
We can observe any periodical function. Most commonly used periodical functions are the trigonometric functions. Thus, we can either take $a_n=\sin(n)$ or $a_n=\cos(n)$. There can be more complex functions, bus since we only need one counterexample, we can stick with those simple ones. For those two sequences there is not a limit $\displaystyle\lim_{n\to\infty}a_n$ since those two will periodically take values from the interval $[-1,1]$ and the terms of the sequence will not approach any value as $n\to\infty$
