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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 $ .

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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.

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