Let $(a_n)$,$(b_n)$ be sequences s.t. $(a_n)\rightarrow l \in \mathbb{R}$ and $(b_n)\rightarrow 0$. Show that $a_nb_n \rightarrow 0$ without using the Limit Laws.

Here's my attempt:

Fix $\epsilon > 0$.

Since $(a_n) \rightarrow l$, $\exists N_1 \in \mathbb{N}$ s.t. $\forall n \geq N_1$, $|a_n-l|< |l|$

Likewise since $(b_n) \rightarrow 0$, $\exists N_2 \in \mathbb{N}$ s.t. $\forall n \geq N_2$, $|b_n|<\frac{\epsilon}{2|l|}$

But $|a_nb_n|=|(a_n-l)b_n + lb_n| \leq |(π‘Ž_π‘›βˆ’π‘™)𝑏_𝑛|+|lb_n| = |a_n-l||b_n|+|l||b_n| < |l| \cdot \frac{\epsilon}{2|l|} +|l| \cdot \frac{\epsilon}{2|l|} = \epsilon$

Since $\epsilon$ was arbitrary, we've shown what's required.

  • 3
    $\begingroup$ Almost there, but you divided by $\lvert l\rvert$ without first dealing with the case $l=0$. Easy fix is to make $\lvert b_n\rvert<\epsilon/(2\lvert l\rvert+1)$ instead, and similarly changing the bound on $a_n-l$. $\endgroup$ – user10354138 Jun 5 '19 at 5:24
  • 2
    $\begingroup$ And you forgot to take/mention $n=\max\{N_1,N_2\}$ when you showed $a_nb_n\to0$ $\endgroup$ – Shubham Johri Jun 5 '19 at 5:26
  • 1
    $\begingroup$ Also your statement $|a_n-l| < |l|$ will not hold if $l=0$. $\endgroup$ – Anurag A Jun 5 '19 at 5:26
  • $\begingroup$ @user10354138, In the case of $|l|=0$ we just make both sequences $\sqrt{\epsilon}$ close to $0$ and take $max\{N_1,N_2\}$ (as I should have mentioned here also), right? $\endgroup$ – alwaysiamcaesar Jun 5 '19 at 5:30

If $l=0$ then we cannot divide by it and the argument does not make sense.

Instead, $a_n$ converges implies it is bounded, say by $M$. Then use $\frac{\varepsilon}{M}$ argument to conclude the result!


Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.