I'm having trouble looking at two problems dealing with sequence limits in advanced calculus. here they are:

A: Suppose that the sequence $\displaystyle a_n$ converges to a nonzero constant $\displaystyle A$. Prove that if $\displaystyle n \geq n*$, we have $\displaystyle |a_n| \geq t|A|$, where t is a fixed real number satisfying $\displaystyle 0 < t < 1$. Is this statement true if $\displaystyle t = 0$? How about $\displaystyle t = 1$ Explain.

I have looked at if $\displaystyle t=0$, it is not true because $\displaystyle 0 |A| \not= 0$, and if $\displaystyle t=1$, then it is true, $\displaystyle 1 |A| \geq 0$. writing the proof has me confused.

B: Prove that if a sequence $\displaystyle a_n$ converges to 0, and a sequence $\displaystyle b_n$ is bounded, then the sequence $\displaystyle a_n b_n$ converges to 0.

I have looked at using the theorem that states if $\displaystyle \lim _{x\to \infty }a_n b_n = A B$, but it requires that $\displaystyle b_n$converges to $\displaystyle B$, what other methods should I use?

2. It seems to me that your part A is confusing.
Observe that if $\displaystyle \left( {a_n } \right) \to A$ then $\displaystyle \left( {\left| {a_n } \right|} \right) \to \left| A \right|.$
There are two cases.
First, if |A|=0 then it is trivially true.
Secondly, if |A|>0 then for $\displaystyle 0 < t < 1$ then let $\displaystyle \varepsilon = \left( {1 - t} \right)\left| A \right|.$
Use the definition: $\displaystyle \left[ {\exists N} \right]\left( {n \ge N \Rightarrow \left| {\left| {a_n } \right| - \left| A \right|} \right| < \varepsilon } \right).$
Thus, $\displaystyle - \left( {1 - t} \right)\left| A \right| < \left| {a_n } \right| - \left| A \right|.$
Can you finish part A?

For part B, use the following as a start:
$\displaystyle \left[ {\exists B > 0} \right]\left( {\left| {b_n } \right| < B} \right)\quad \& \quad \varepsilon = \frac{\delta }{B}$

3. I have just finished my Advance Calculus exam and just drop by to see what's up.

Now I have a question on Plato's advise on part B, how do you know the sequence of Bn is converging to B? It is bounded, but it doesn't have to converge.

Sad to say, I'm still trying to figure out how to solve the problems even thou I completed the course...

KK

The whole point is that we do not know that $\displaystyle {b_n }$ converges!
In fact, $\displaystyle {b_n }$ may not converge at all!
The only claim is that $\displaystyle {b_n }$ is bounded by B.