# Proving a^b Convergence from a_n and b_n Limits

• daniel_i_l
In summary: Yes, that is correct. In summary, the proof involves splitting a_n into two subsequences based on whether a_n is greater than or equal to 1 or less than 1. By using the sandwich rule and the fact that logarithmic and exponential functions are continuous, it can be shown that the limit of (a_n)^(b_n) is equal to a^b.
daniel_i_l
Gold Member

## 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).

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

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.

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.

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.

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

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.

anyone?
Thanks.

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

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.

daniel_i_l said:
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.

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

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.

daniel_i_l said:
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.

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

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

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.

## What is meant by "proving a^b convergence from a_n and b_n limits?"

"Proving a^b convergence from a_n and b_n limits" refers to using the limits of two sequences, a_n and b_n, to show that the sequence a^b converges. This means that as n approaches infinity, the values of a^b get closer and closer to a single, finite number.

## Why is it important to prove a^b convergence from a_n and b_n limits?

Proving a^b convergence from a_n and b_n limits is important because it allows us to determine the behavior of the sequence a^b without having to directly compute its limit. This can be useful in many applications, such as in calculus and physics, where finding the limit of a complicated sequence may be challenging or impossible.

## What is the general method for proving a^b convergence from a_n and b_n limits?

The general method for proving a^b convergence from a_n and b_n limits is to use the limit laws and properties to manipulate the expression a^b into a form that can be evaluated using the limits of a_n and b_n. This often involves using algebraic manipulations and theorems such as the squeeze theorem.

## What are some common challenges when proving a^b convergence from a_n and b_n limits?

Some common challenges when proving a^b convergence from a_n and b_n limits include the complexity of the expressions involved, the necessity of using various limit laws and properties, and the need for careful algebraic manipulations. Additionally, it can be difficult to determine the correct approach for a specific problem and to ensure that all steps are logically sound.

## How can one improve their skills in proving a^b convergence from a_n and b_n limits?

One can improve their skills in proving a^b convergence from a_n and b_n limits by practicing with a variety of examples and problems, familiarizing themselves with the limit laws and properties, and seeking help from resources such as textbooks, online tutorials, and tutoring sessions. It can also be helpful to review and understand theorems related to limits, such as the squeeze theorem and the limit laws.

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