I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.
The Prompt:
Show that if $\sum{x_n}$ converges and $(y_n)$ is bounded and monotone (either increasing or decreasing), then $\sum{x_ny_n}$ converges.
My Proof:
Suppose $\sum{x_n}$ converges, so that for all $ε>0$ there exists a $n_ε$ such that $|\sum_n^m{x_n}| \le{ε}$ for all $m \le{n} \le{n_ε}$. If $sup |y_n|=0$ then $(y_n)$ is a sequence of all zeros and so $\sum{x_ny_n}=0$ and thus converges trivially. Now suppose $sup|y_n| \neq 0$ and define $ε_y = ε * sup|y_n|$. Since $(y_n)$ is bounded, $sup |y_n|$ is finite and so $ε_y$ is finite. Also, since $ε$ can be thought of as any element of $\{x|x>0\}$ and since $sup|y_n|$ is a positive constant, $ε_y ∈ \{x|x>0\}$ as $\{x|x>0\} = \{x|cx>0\}$ for any positive constant $c$. Then for every $ε_y>0$ there exists a $n_ε$ (the same $n_ε$ as for $\sum_n^m{x_n} \le{ε}$) such that:
$\sum_n^m{x_ny_n} \le{sup|y_n| * \sum_n^m{x_n}} \le{sup|y_n| * ε} = ε_y$
holds for all $m \le{n} \le{n_ε}$ and thus, $\sum{x_ny_n}$ is convergent.
*I didn't use the fact that $(y_n)$ was monotone in my proof. Am I missing something?
Your proof is wrong. Actually, if you exclude the assumption that $(y_n)_{n\in\Bbb N}$ is monotonic, the statement is false, and so it cannot be proved. You were supposed to prove that $\bigl\lvert\sum_{k=m}^nx_ky_k\bigr\rvert<\varepsilon$, for each $\varepsilon>0$. You replaced $\varepsilon$ with another number and you forgot the absolute value.
By the way, what you are trying to proof is known as Abel's test.