# If $\{x_n\}$ is a sequence that diverges, then $\{cx_n\}$ diverges also for $c\ne 0$

PROOF: Suppose that $\{cx_n\}$ is, in fact, convergent to some limit $L$. Then for $\epsilon = |c| > 0$ there exists $N \in \mathbb N$ such that for all $n\ge N$, we have $$|cx_n - L | < |c| \implies |cx_n| - |L| < |c| \implies |cx_n| < |c| + |L| \implies |x_n| < 1 + \left|{L \over c}\right|,$$

which shows that $\{x_n\}$ is bounded.

Here's where I'm stuck. If $\{x_n\}$ diverges to $\pm\infty$, then we are done since we've reached a contradiction. However, I'm not sure what we could do here for if $\{x_n\}$ oscillates. Obviously this argument won't work because "oscillates" could apply to either bounded or unbounded sequences. Can someone provide a hint on what we know here that helps us conclude that an oscillating sequence would diverge also?

You're basically trying to show that if $cx_n \to L$, then $x_n \to \frac{L}{c}$. To this end, choose any $\epsilon > 0$. Then since $cx_n \to L$, we know that for $\epsilon' = |c|\epsilon$ (which is certainly positive), there is some $N' \in \mathbb N$ such that if $n \geq N'$, then $|cx_n - L| < \epsilon' = |c|\epsilon$. So if we choose $N = N'$, it follows that if $n \geq N$, then: $$\left|x_n - \frac{L}{c} \right| = \left| \frac{cx_n - L}{c} \right| = \frac{1}{|c|}|cx_n - L| < \frac{1}{|c|}(|c|\epsilon) = \epsilon$$ as desired.
$\{x_n\}$ converges $\Leftrightarrow$ $\forall \epsilon > 0 \exists n_0 : \forall n,m > n_0 |x_m - x_n| < \epsilon$.
Then $\{x_n\}$ diverges $\Leftrightarrow$ $\forall \epsilon > 0 \ \forall n_0 : \exists n,m > n_0: |x_m - x_n| > \epsilon$.
Let $\epsilon=\epsilon_0c^{-1}$. So since $\{x_n\}$ diverges $|x_m - x_n| > c^{-1}\epsilon_0$. $|cx_m - cx_n| > \epsilon_0$ for all $n_0$ and each $\epsilon=\epsilon_0$. So ${cx_n}$ diverges.