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

Use reductio ad absurdum.

19. Sep 28, 2010

eibon

please explain how you would do the contradiction

20. Sep 28, 2010

Dickfore

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?