Prove: If the sequence $<a_{n}>$ converges to $0$, and the sequence $<b_{n}>$ is bounded, then the sequence $<a_{n}b_{n}>$ also converges to $0$.
Let $\epsilon > 0$, since $\lim_{n \to \infty}<a_{n}> = 0$ applying the definition of the limit $$|a_{n}| = |a_{n} - 0| < \epsilon \ \ \forall n\in\mathbb{N}$$ Then, $$|a_{n}b_{n} - 0| = |a_{n}b_{n}| < \epsilon \ \ \forall n\in\mathbb{N}$$
I am not sure if I am right, any suggestions is greatly appreciated
The strategy for this proof is this: since $a_n \to 0$ we can make $a_n$ arbitrarily small and since $b_n$ is bounded we know that there is a bound on the size of $|b_n|$. Hence, we will make $|a_n|$ small enough so that $|a_n b_n|$ is always less than $\epsilon$. Here is a proof:
Let $\epsilon>0$. Since $b_n$ is bounded then there exists $M\in\mathbb{R}$ such that $$ |b_n| < M $$ for all $n > 0$. Since $a_n\to0$ then we can choose $N\in\mathbb{N}$ such that $|a_n| < \epsilon/M$ for all $n \ge N$. Hence, for all $n\ge N$ we have $$ |a_n b_n| < M|a_n| < M\cdot \frac{\epsilon}{M} = \epsilon.$$
Therefore, $|a_n b_n| \to 0$.
- not for all $n\in\mathbb N$, it must be changed with "sufficiently large $n$'s" 2. $\epsilon M$. But your proof is nearly correct.
– MBYagbasan Feb 20 '15 at 17:53