# What is the correct definition of $a_n\to-\infty$?

My instructor notes (his written notes on blackboard while teaching) states that

The sequence $a_n\to-\infty$ as $n\to\infty$ if for all $M\in\mathbb{R}$, there exists $N\in\mathbb{N}$ such that $$n\geq N\Rightarrow a_n<M$$

However, when I try to show that the sequence $a_n=n-n^2$ approaches $-\infty$ as $n\to\infty$, I have difficulty because I am not sure what $N$ needs to be.

Is it likely that there is a typo in the definition and the definition should really be the following?

The sequence $a_n\to-\infty$ as $n\to\infty$ if for all real $M<0$, there exists $N\in\mathbb{N}$ such that $$n\geq N\Rightarrow a_n<M$$

If the second definition is right, how can I use this second definition to show the divergence of the sequence $a_n=n-n^2$?

This is my analysis so far: When I try to solve the inequality $n-n^2<M$ which is equivalent to $n^2-n+M>0$, my discriminant of the quadratic in $n$ is $1-4M$, so now I have two consider three cases: (1) If $M=1/4$, then $n^2-n+M$ is a perfect square and so $n^2-n+m>0$ is true for all $n\geq1$. (2) If $1-4M<0$, then $n^2-n+M$ does not cut the $x$-axis and so is always positive since it is above the $x$ axis for all $n\geq1$. (3) If $1-4M>0$, then I get two roots and $n^2-n+M>0$ for all $n<n_1$ or $n>n_2$ where $n_1$ and $n_2$ are roots of the quadratic equation $n^2-n+M$. So if I take $N\geq\max\{1,n_2\}$, then it seems to work but I am not sure if it is correct.

• Those two definitions are equivalent. See this question: math.stackexchange.com/questions/1782295/… Commented Mar 17, 2017 at 6:29
• The immediate assumption that the teacher is wrong is worth pondering.
– Did
Commented Mar 17, 2017 at 6:32
• When I try to solve the inequality $n-n^2<M$ which is equivalent to $n^2-n+M>0$, my discriminant of the quadratic in $n$ is $1-4M$, so now I have two consider three cases: (1) If $M=1/4$, then $n^2-n+M$ is a perfect square and so $n^2-n+m>0$ is true for all $n\geq1$. (2) If $1-4M<0$, then $n^2-n+M$ does not cut the $x$-axis and so is always positive since it is above the $x$ axis for all $n\geq1$. (3) If $1-4M>0$, then I get two roots and $n^2-n+M>0$ for all $n<n_1$ or $n>n_2$ where $n_1$ and $n_2$ are roots of the quadratic equation $n^2-n+M$.
– user415849
Commented Mar 17, 2017 at 6:36

If we are to prove $a_{n} \to -\infty$, then considering those bounds $<0$ suffices. Likewise if we are to prove $a_{n} \to +\infty$, then checking those bounds $>0$ suffices.
Note that $n-n^{2} = n(1-n)$ for all $n$. Note that $n \geq 2$ implies $1-n \leq -1$, implying that $n(1-n) \leq -n$, which is $<M$ if in addition $n \geq -M$. So $N := \max \{ 2, \lceil -M \rceil +1 \}$ suffices.
The point is to bound away the annoying $1-n$ by using a convenient preliminary bound for $n$.