# For $0 \leq a_n \leq b_n$ for every n and $\{b_n\} = 0$ Show $\{a_n\} \rightarrow 0$

If $\{a_n\}$ and $\{b_n\}$ are sequences with $0 \leq a_n \leq b_n$ for every n and $\{b_n\} \rightarrow 0$ show that $\{a_n\} \rightarrow 0$

need help completing proof by contradiction:

Suppose $\{a_n\} \rightarrow 1$. Pick $\epsilon =$ ? . Since $\{b_n\} \rightarrow 0$

we have, $|a_n -1| < \epsilon$ and $|b_n - 0| < \epsilon$ Since $0 \leq a_n \leq b_n, |b_n - a_n|<$

• Why are you assuming $a_n \to 1$? – user99914 Oct 8 '15 at 8:25
• Why you suppose $a_n \rightarrow 1$ ?? – Nizar Oct 8 '15 at 8:26
• I did this to contradict that it converges to 0 – David Oct 8 '15 at 8:26
• Even if you show $a_n$ does not converge to $1$, it might still not converging to $0$. – user99914 Oct 8 '15 at 8:27

If you want to do a proof by contradiction, you need to assume the contrary of $a_n \to 0$, which is not $a_n \to 1$, but $a_n \not\to 0$, that is$\def\eps{\varepsilon}$ $$\exists \eps > 0 \;\forall N \; \exists n \ge N: |a_n| \ge \eps.$$ But here a direct proof works fine: (Hint, I'm sure you can fill in the steps) Let $\eps > 0$ we want to find an $N$ such that $|a_n| < \eps$ for $n \ge N$. But $b_n \to 0$. Choose $N$ according to that, now use $0 \le a_n \le b_n$.