True or false (prove if true):

  1. If $\{a_n\}$ converges to $l\neq 0$ and $\{b_n\}$ diverges, then $\{a_nb_n\}$ diverges.
  2. If $\{\frac{a_n}{b_n}\}$ converges to an irrational number $l$, then that $|b_n|\rightarrow \infty$ as $n\rightarrow \infty$.


  1. I know if $a_n=1/n$ and $b_n=n$ then $\{a_nb_n\}$ converges to 0. But here, $a_n\rightarrow l\neq 0$. Please provide an answer.

    1. $\{\frac{a_n}{b_n}\}$ converges to an irrational number $l$ then $$|\frac{a_n}{b_n}-l|<\epsilon$$ i.e $$|a_n-lb_n|<\epsilon |b_n|$$. Nothing is given about $\{a_n\}$. I tried to prove by contradiction by assuming $b_n$ is bounded but fails. Please help.
  • 1
    $\begingroup$ The first one is true, no use looking for a counter example. The second one is false, no use trying to prove it. Hints: 1 Suppose $(a_nb_n)_{n\in \mathbb N}$ converges. Consider the sequence $\left(\frac{1}{a_{n+k}}\right)_{n\in \mathbb N}$ (where $k$ is a natural number such that $\forall m\in \mathbb N(k\leq m\implies a_m\neq 0)$ - why does such a $k$ exist?). 2 Let the denominator be a rational constant. $\endgroup$ – Git Gud May 9 '16 at 20:25
  • $\begingroup$ For the last one, are we to assume that $a_n,b_n\in \mathbb N$? $\endgroup$ – lulu May 9 '16 at 20:28
  1. Yes, $\{a_nb_n\}$ diverges. For assume $a_nb_n\to L$. Then $b_n=\frac{a_nb_n}{a_n}\to \frac Ll$, contradiction

  2. No. You might have $a_n=\pi$ and $b_n=1$ for all $n$, so $\frac{a_n}{b_n}\to \pi$ and $b_n$ bounded.

  • $\begingroup$ @Liad Sorry, you aint right. $\endgroup$ – zhw. May 9 '16 at 20:34

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