# Sequence limit - real analysis

1. Sep 27, 2007

### antiemptyv

1. The problem statement, all variables and given/known data

Prove that the sequence $$(x_n) = ((a^n+b^n)^{1/n})$$ converges to $$b$$, for 0 < a < b.

3. The attempt at a solution

I haven't dealt with any sequences with $$n$$'s in the exponent, but I assume I'll have to use logarithms at some point to get at them? Can someone start me off in the right direction?

Last edited: Sep 27, 2007
2. Sep 27, 2007

### Hurkyl

Staff Emeritus
The whole idea is that a introduces an increasingly insignificant error. Therefore, differential calculus should be applicable. Personally, I would try a differential approximation.

Last edited: Sep 27, 2007
3. Sep 27, 2007

### antiemptyv

here's a follow-up attempt

What about something like $$|(a^n+b^n)^{1/n}-b| < (2b^n)^{1/n}-b < 2^{1/n} < 2^{1/\epsilon}$$ when $$n > \epsilon$$?

4. Sep 27, 2007

### Hurkyl

Staff Emeritus
Why is $(2b^n)^{1/n} - b < 2^{1/n}$?

And more importantly, why would $|(a^n + b^n)^{1/n} - b| < 2^{1/\epsilon}$ be helpful?

Last edited: Sep 27, 2007
5. Sep 27, 2007

### antiemptyv

i think i meant...

$$(2b^n)^{1/n}- b = 2^{1/n}b - b < 2^{1/n}b$$

sorry, i think it's just because it's late.

6. Sep 27, 2007

### Hurkyl

Staff Emeritus
See the new question I added.

7. Sep 27, 2007

### antiemptyv

if $$n>\epsilon$$, then $$\frac{1}{n} < \frac{1}{\epsilon}$$, and $$2^{\frac{1}{n}} < 2^{\frac{1}{\epsilon}}$$

Last edited: Sep 27, 2007
8. Sep 27, 2007

### VietDao29

Well, and how can you go from there?

Remember that:
$$\lim_{\epsilon \rightarrow + \infty} 2 ^ {\frac{1}{\epsilon}} = 1$$, not 0.

How about taking bn outside, like this:

$$\lim_{n \rightarrow \infty} \left\{ b ^ n \left[ \left( \frac{a}{b} \right) ^ n + 1 \right] \right\} ^ \frac{1}{n}$$

Can you go from here? :)