How do I symbolically prove that $\lim_{n \to \infty } (n-n^2)=- \infty $?

Question

Intuitively we know that $n^2$ grows faster than $n$, thus the difference tends to negative infinity. But I have trouble proving it symbolically because of the indeterminate form $\infty - \infty$. Is there anyway to do this without resorting to the Epsilon-Delta definition ?

Answer 1

Note that when $n\ge 2$,

$$n - n^2 \le n \left(\frac n2\right) - n^2 = -\frac{n^2}{2},$$

as $$\lim_{n\to \infty} -\frac{n^2}{2} = -\infty,$$

then so are $\lim (n-n^2)$.

Answer 2

Another way: complete the square $n-n^2=\frac14-\bigl(n-\frac12\bigr)^2\to -\infty$.

Answer 3

Let $M < 0$; then $n-n^{2} = n(1-n) < n(1-5) = -4n < M$ if $n > \max \{ 5, |M|/4 \}$.