Proving a^b Convergence from a_n and b_n Limits

[daniel_i_l]

Homework Statement

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).

Homework Equations

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}} >= 1and 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+Eand 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.

 

[e(ho0n3]

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

 

[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.

 

[daniel_i_l]

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

 

[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.

 

[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

 

[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.

 

[daniel_i_l]

anyone?
Thanks.

 

[HallsofIvy]

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

 

[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} &lt; a_i^{b_i} &lt; (a + E)^{b + E}. Of course, what I really need to show is that a^b - E &lt; a_i^{b_i} &lt; a^b + E. This is where I'm stumped.

 

[NateTG]

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.

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.

 

[daniel_i_l]

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

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

 

[HallsofIvy]

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

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.

 

[daniel_i_l]

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