# Thread: Help with Proofs

1. ## Help with Proofs

I'm taking a new class and instructor isn't very helpful . These are probably very simple, but I'm lost.

1) If an-->L, then |an|-->|L|

2) Prove that lim(an-bn) = lim(an)-lim(bn), provided lim(an) and lim(bn) exist

3) Give an example where lim(an) and lim(bn) do not exist, but lim(an+bn) exists.

4) If {an} is a bounded sequence and if {bn} is a sequence converging to 0, then {anbn} converges to 0.

I have tons of questions, but I'll start with these.

Help is extremely appreciated!

2. Originally Posted by abradfo1
I'm taking a new class and instructor isn't very helpful . These are probably very simple, but I'm lost.

1) If an-->L, then |an|-->|L|

2) Prove that lim(an-bn) = lim(an)-lim(bn), provided lim(an) and lim(bn) exist

3) Give an example where lim(an) and lim(bn) do not exist, but lim(an+bn) exists.

4) If {an} is a bounded sequence and if {bn} is a sequence converging to 0, then {anbn} converges to 0.
The secret to #1 is $\displaystyle \left| {\left| {a_n } \right| - \left| L \right|} \right| \leqslant \left| {a_n - L} \right|$.

For #4 Suppose that $\displaystyle \left( {\forall n} \right)\left[ {\left| {a_n } \right| \leqslant A} \right]$ then $\displaystyle \left| {a_n b_n } \right| = \left| {a_n } \right|\left| {b_n } \right| \leqslant A\left| {b_n } \right|$

3. Originally Posted by abradfo1
I'm taking a new class and instructor isn't very helpful . These are probably very simple, but I'm lost.

1) If an-->L, then |an|-->|L|

2) Prove that lim(an-bn) = lim(an)-lim(bn), provided lim(an) and lim(bn) exist

3) Give an example where lim(an) and lim(bn) do not exist, but lim(an+bn) exists.

4) If {an} is a bounded sequence and if {bn} is a sequence converging to 0, then {anbn} converges to 0.

I have tons of questions, but I'll start with these.

Help is extremely appreciated!
2. Since $\displaystyle \{a_n\}\to a$, $\displaystyle \exists~N_1$ such that $\displaystyle n>N_1$ implies that $\displaystyle |a_n-a|<\frac{\epsilon}{2}$. Similarly, since $\displaystyle \{b_n\} \to b$, $\displaystyle \exists~N_2$ such that $\displaystyle n>N_2$ implies that $\displaystyle |b_n-b|<\frac{\epsilon}{2}$.

Let $\displaystyle N=\max\{N_1,N_2\}$. Thus, when $\displaystyle n>N, |(a_n-b_n)-(a-b)| = \underbrace{|(a_n-a)+(b-b_n)| \leq |a_n-a|+|b-b_n|}_{triangle~inequality} < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$.

Therefore $\displaystyle \lim(a_n-b_n)=a-b = \lim a_n-\lim b_n$.

3. Consider $\displaystyle a_n = 1+(-1)^n$ and $\displaystyle b_n=1-(-1)^n$.

4. Thanks so much. I just had no idea where to start! Thank you!