Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working on this problem but can't find an insight about how to proceed.

The statement: Let $0 \lt a \le b$ and $(x_n) = (a^n + b^n)^\frac1n$ then $ \lim_{n \rightarrow \infty} (x_n) = b$

In an example in the textbook prove that $ (c^\frac1n) \rightarrow 1$ using the expression $c^\frac1n = 1 + d_n$ where $d_n \gt 0$ and then the Bernoulli inequality but does not explain well why there there is $d_n$.

Any hint will be appreciated.

UPDATE

I come up with this regarding $2^\frac1nb$:

Fixing $\epsilon \gt 0$ that $2^\frac1nb \lt b + \epsilon$ then $2 \lt (1 + \frac{\epsilon}b)^n$. By Bernoulli inequality we have $(1 + \frac{\epsilon}b)^n \ge 1 + n\frac{\epsilon}b$. We have that $1 < n\frac{\epsilon}b$ when $n \gt \frac{b}{\epsilon}$. Finally for $n \gt N = [\frac{b}{\epsilon}]$ it follows $2 < 1 + n\frac{\epsilon}{b} \le (1 + \frac{\epsilon}{b})^n$ thus $2^\frac1nb \lt b + \epsilon$. Thank you all for your hints.

share|improve this question
add comment

3 Answers

up vote 9 down vote accepted

Hint: Use that

$$b^n \leq a^n + b^n \leq 2b^n.$$

share|improve this answer
    
+1. Much much simpler than what I have written! –  user17762 Oct 19 '11 at 1:09
    
Cheers. We can just consider your hint more sophisticated :) –  JavaMan Oct 19 '11 at 1:29
    
Awesome, but it seems like it should be $b^n \lt a^n + b^n \le 2b^n$ as $a^n \gt 0$, right? –  Rho Oct 19 '11 at 2:28
    
What I said is true, though strictly speaking, you are correct. Just be careful in taking limits as it is tempting to say that $b < \lim \leq b$, which is incorrect! –  JavaMan Oct 19 '11 at 2:33
add comment

HINT:

First note that $x_n > b$, $\forall n \in \mathbb{N}$ (Why?).

Now show that $b$ is the greatest lower bound for the sequence $\{x_n\}_{n=1}^{\infty}$.

To be more specific, show that given any $\epsilon > 0$, $\exists N \in \mathbb{N}$ such that $\forall n > N$, we have $a^n + b^n < (b+\epsilon)^n$.

For instance, choose $N = \left \lceil \frac{a}{\epsilon} \right \rceil$ and expand by binomial theorem to argue out.

share|improve this answer
    
Thank you very much. I'm going to work on this. –  Rho Oct 19 '11 at 1:00
add comment

If you factor $b^n$ from the expression in parentheses, you can apply the Bernoulli inequality to the remaining expression and get the result you're hoping for.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.