WTS:
(1) $\exists a \in \mathbb R, \forall \epsilon > 0, \exists N_1 > 0$, such that for all $n \in \mathbb N$, if $n > N_1$, then $|a_n – a| <$ __
(2) $\exists b \in \mathbb R, \forall \epsilon > 0, \exists N_2 > 0$, such that for all $n \in \mathbb N$, if $n > N_2$, then $|b_n – b| < \epsilon$
Let $\epsilon > 0$ is arbitrary
Choose $N = $ ____ > 0
Let $b_n = \frac{a_nb_n}{a_n}$
Suppose $n > N$, then
$$|\frac{a_nb_n}{a_n} – \frac{ab}{a}| = |\frac{a_nb_na – aba_n}{a_n a}| = |\frac{a_na(b_n – b)}{a_n a}| = |b_n – b| < \epsilon$$
Yeah. I don't think its right.. (plus a = 0)
I know how to solve this if it was If $\{a_n\}$ and $\{b_n\}$ are convergent sequences, then $\{a_nb_n\}$ converges. But not what the question gave.
Can anyone point me to the right direction? Thx
Best Answer
Let $a_{n} := 1/n^{2}$ for all $n$; let $b_{n} := n$ for all $n$. Then $(a_{n})$ converges and $(a_{n}b_{n}) = (1/n)$ still converges. But $(b_{n})$ ain't.