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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.


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.

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up vote 10 down vote accepted

Hint: Use that

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

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+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? – Rolando 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


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.

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Thank you very much. I'm going to work on this. – Rolando Oct 19 '11 at 1:00

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.

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