Homework Help: Series question

1. Feb 28, 2008

daniel_i_l

1. The problem statement, all variables and given/known data
$$a_{n} -> a$$ , a>0 and $$a_{n}>0$$ for all n. Prove that if $$b_{n} -> b$$ then $${(a_{n})}^{(b_{n})} -> a^b$$.
(-> means the limit as n goes to infinity).

2. Relevant equations

3. The attempt at a solution

I first split up a_n into to sub series: $$a_{n_{k}}$$ and $$a_{n_{j}}$$ where for all k $$a_{n_{k}} >= 1$$and for all j $$0<a_{n_{j}}<1$$ . Now, starting with the first series, for all E>0 we can find a number N so that for all k>N $$b-E < b_{n_{k}} < b+E$$and so
$${(a_{n_{k}})}^{(b-E)} < {(a_{n_{k}})}^{(b_{n_{k}})} < {(a_{n_{k}})}^{(b+E)}$$. I know that $${(a_{n_{k}})}^{(b+E)} -> a^{(b+E)}$$ and so that since E can be as small as we want we should be able to use the sandwich rule to prove that $${(a_{n_{k}})}^{(b_{n_{k}})} -> a^b$$ and we can do the same to prove that $${(a_{n_{j}})}^{(b_{n_{j}})} -> a^b$$ . But how can I write it in a rigorous way?
Thanks.

Last edited: Feb 28, 2008
2. Feb 28, 2008

e(ho0n3

What's the point of splitting a_n into two subsequences?

3. Feb 28, 2008

e(ho0n3

If the limit as n goes to infinity of (a_n)^(b_n) is a^b, then for any E > 0, there exists an N such that for all indices i >= N, |(a_i)^(b_i) - a^b| < E. To demonstrate this (rigorously), you will need to tell me what this N is.

4. Feb 28, 2008

daniel_i_l

I split it in two so that I could seperate between cases where a>=1 and a<1. Can you give me a hint how to find N?
Thanks.

5. Feb 28, 2008

e(ho0n3

Does it matter whether a < 1 or a >= 1? I don't get it.

You know two things:

- For all E > 0, there is an N_a such that for all i > N_a, |a_i - a| < E.
- For all E > 0, there is an N_b such that for all i > N_b, |b_i - b| < E.

Obviously the N will have something to do with N_a and N_b. Determine what that something is.

6. Feb 28, 2008

Mystic998

Well, if you don't mind a cheat-ish answer, you could just look at the logarithm, then use the facts that logarithmic and exponential functions are continuous and that products of convergent sequences converge to the product of the limits

7. Feb 28, 2008

daniel_i_l

If a_i < a+E and b_i < b+E you still can't write (a_i)^(b_i) < (a+E)^(b+E) since you don't know whether a+E is bigger that 1 or not. That's what's confusing me.

8. Feb 29, 2008

daniel_i_l

anyone?
Thanks.

9. Feb 29, 2008

HallsofIvy

What can you base your proof on? If you are allowed to use the fact that ax is continuous, it's almost trivial.

10. Feb 29, 2008

e(ho0n3

What I was thinking is the following: Let N be the greater of $N_a$ and $N_b$. Then surely $(a - E)^{b - E} < a_i^{b_i} < (a + E)^{b + E}$. Of course, what I really need to show is that $a^b - E < a_i^{b_i} < a^b + E$. This is where I'm stumped.

11. Feb 29, 2008

NateTG

You can take the absolute value - you're trying to shrink:
$$|a^b-a_n^{b_n}|$$
so it doesn't matter which way the inequality goes.

12. Mar 1, 2008

daniel_i_l

How is it trivial? I know that $${a}^{b_{n}} -> a^b$$ but how do I prove that $${a_{n}}^{b_{n}} -> a^b$$
Thanks.

13. Mar 1, 2008

HallsofIvy

For every m, $a_m^{b_n}\rightarrow a_m^b$. Then $a_m^b\rightarrow a^b$.

It follows from that that $a_n^{b_n}\rightarrow a^b$.

14. Mar 2, 2008

daniel_i_l

Thanks, is this what you meant:
for all m $${a_m}^{b_n} -> {a_m}^b$$so for every E there's some N so that for all n>N $${a_m}^b -E < {a_m}^{b_n} < {a_m}^b + E$$ and since $${a_m}^{b} -> {a}^b$$so for every F there's some M so that for all m>M $${a}^b -F < {a_m}^{b} < {a}^b + F$$ and so if P>max{M,N} then for all m,n > P
$${a_m}^b -(E+F) < {a_m}^{b_n} < {a}^b + (E+F)$$ QED
Is that right?
Thanks.