# Homework Help: Hard sequence queston (calculus)

1. Sep 27, 2010

### eibon

1. The problem statement, all variables and given/known data
if {An} converges and lim {An}=L L$$\neq$$ 0
and {Bn} diverges then does {An X Nn} diverge? prove formally

2. Relevant equations

3. The attempt at a solution

can anyone give me a hint or show a solution?

2. Sep 28, 2010

### hunt_mat

Examine the sequences:
$$a_{n}=1-\frac{1}{\sqrt{n}},\quad b_{n}=n$$

3. Sep 28, 2010

### eibon

i dont see how that helps

4. Sep 28, 2010

### Dick

You'll probably find it easier to organize your proof as a proof by contradiction. Assume {a_n*b_n} converges, say to M, and a_n converges to a nonzero L. Can you show that means b_n would converge to M/L?

5. Sep 28, 2010

### eibon

yeah i thought of proof by contradiction to day, but i end up getting (M/(L+$$\epsilon$$1)) < Bn< (m+$$\epsilon$$2)/L

and how do you get it converges to M/L? do i need to let epsilon = something?

Last edited: Sep 28, 2010
6. Sep 28, 2010

### eibon

L<An<L +$$\epsilon$$ $$\leftrightarrow$$ $$\frac{1}{L +epsilon}$$ <$$\frac{1}{An}$$ < $$\frac{1}{L}$$

and M<AnBn<M+ε

then i get $$\frac{M}{L+epsilon}$$ <Bn< $$\frac{M+epsilon}{L}$$

now what do i do?

Last edited: Sep 28, 2010
7. Sep 28, 2010

### Dick

You could stop being so sloppy for one thing. What you actually want to say is for every e>0 there is an N such that for all n>N, L-e<=An<=L+e. Now if you pick e small enough so that L-e>0 (assuming L>0) then you can say 1/(L+e)<=1/An<=1/(L-e). Same sort of thing for An*Bn. Finally you want to show that you can make |Bn-M/L| as small as you want by picking e small enough. You could skip all of this epsilon-delta business if you have a theorem that says if an->A and bn->B and B is nonzero then an/bn->A/B.

8. Sep 28, 2010

### eibon

wait how do you get to |Bn-M/L|<e?

9. Sep 28, 2010

### Dick

You write down an expression like ...<Bn<... and subtract M/L. Then take the absolute value and show you can make the upper and lower bounds as small as you want. Are you sure you don't have a theorem about the quotient of limits?

10. Sep 28, 2010

### eibon

so some thing like this?
and for the quotient of limits you need both An and Bn to converge to use that, but Bn does not converge so i cant use that

Last edited: Sep 28, 2010
11. Sep 28, 2010

### Dick

No. For one thing |an-L|<e becomes L-e<an<L+e. It doesn't imply L<an. And choosing your epsilons to be L and M doesn't help at all. One more time with the question, you don't have a theorem about lim an/bn when lim bn is not zero?

12. Sep 28, 2010

### eibon

for the quotient of limits you need both An and Bn to converge to use that, but Bn does not converge so i cant use that. and i dont understand how to get the upper and lower bounds, can you please explain it?

13. Sep 28, 2010

### Dick

|An-L|<e means -e<An-L<e. And if you have the limit of quotients theorem, apply it to an*bn/an. The numerator and denominator of that both converge if you are assuming an*bn->M. Applying the theorem will save you from all this epsilon monkey business, which you don't seem to have a very good handle on. No offense.

14. Sep 28, 2010

### eibon

then is this right ?
and that says DNE not one in the picture

15. Sep 28, 2010

### Dick

That's basically it. You should maybe mention you can apply the theorem because lim (an)=L and L is not zero.

16. Sep 28, 2010

### eibon

ok thanks , just out of curiosity how would you prove it with epsilons

17. Sep 28, 2010

### eibon

is this right using epsilons?

18. Sep 28, 2010

### Dickfore

19. Sep 28, 2010

### eibon

1. If $\lim_{n \rightarrow \infty}{\{a_{n}\}} = L, \; L \neq 0$ show that, starting from some $n_{0}$, $|a_{n}| > 0, \forall n \ge n_{0}$;
2. Assume the opposite is true, namely $c_{n} \equiv a_{b} \, b_{n}$ converges;
3. Because of 1, it makes sense to define the sequence $c_{n}/a_{n}, \; \forall n \ge n_{0}$. Because both $\{a_{n}\}$ and $\{c_{n}\}$ are convergent, what can we say about their quotient?