# Prove the theorem: if $(a_{n}) \rightarrow \infty$ and $d < 0$ then $(da_{n}) \rightarrow -\infty$

I've attempted this question and my solution is below, just wanted to check that this is correct and that it makes sense.

Let c < 0 be arbitrary, and let $N = \frac{c}{d}$.

Then $\forall n > N, \ da_{n} < dN \leqslant \frac{c \cdot d}{d} = c \ .$

So $\forall c < 0, \ \exists N \in \mathbb{N}$ such that $\forall n > N, da_{n} < c$ .

Therefore $(da_{n}) \rightarrow -\infty$ .

A rigorous way of showing that a sequence "converges to infinity" is to show that $\forall \epsilon$ $\exists N$ such that $n > N$ implies $a_n > \epsilon$. Intuitively, what this means is that for any number A, I can find a term in $a_n$ past which all terms are greater than A.

So if the sequence "converges to negative infinity" we just use the same $N$ as above, and we multiply by $d < 0$ so we have $d*a_n < \epsilon*d$ (we switch orientation of "<" if we multiply or divide by a negative number) . This implies that the sequence will eventually get smaller than any negative real number, meaning $a_n$ ==> $-\infty$. Hope this helps.