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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

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  • 2
    $\begingroup$ Hint: find a counterexample. You already noted that $a=0$ could be problematic, so it seems a good idea do use this to find the counterexample $\endgroup$ – PhoemueX Mar 22 '17 at 7:34
  • $\begingroup$ Ohhhh, I thought this statement was True $\endgroup$ – user349557 Mar 22 '17 at 7:35
  • $\begingroup$ Hint: let $b_n=(-1)^n$. $\endgroup$ – adfriedman Mar 22 '17 at 7:37
  • $\begingroup$ $a_n = \frac{1}{x}$ and $b_n = x$, is what I came up with actually. Could you tell me your counter example with $b_n = (-1)^n$? I cant come up with one using that. $\endgroup$ – user349557 Mar 22 '17 at 7:40
  • $\begingroup$ Yours works as well. My example was $a_n=\frac{1}{n}$ and $b_n=(-1)^n$, then $a_n \to 0$ and $a_nb_n \to 0$ but $b_n$ doesn't converge. This a non-trivial case that has the added insight that $b_n$ can be bounded and the claim still fails (this differs from an analogous theorem about series). $\endgroup$ – adfriedman Mar 23 '17 at 18:53
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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.

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  • $\begingroup$ Even easier: $a_n = 0$ for all $n$ :) $\endgroup$ – PhoemueX Mar 22 '17 at 8:56
  • $\begingroup$ Haha, true! Yes $\endgroup$ – Megadeth Mar 22 '17 at 8:59

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