Sequence limit - real analysis

[antiemptyv]

Homework Statement

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

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?

 

[Hurkyl]

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

 

[antiemptyv]

here's a follow-up attempt

What about something like |(a^n+b^n)^{1/n}-b| &lt; (2b^n)^{1/n}-b &lt; 2^{1/n} &lt; 2^{1/\epsilon} when n &gt; \epsilon?

 

[Hurkyl]

Why is (2b^n)^{1/n} - b &lt; 2^{1/n}?

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

 

[antiemptyv]

i think i meant...

(2b^n)^{1/n}- b = 2^{1/n}b - b &lt; 2^{1/n}b

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

 

[Hurkyl]

See the new question I added.

 

[antiemptyv]

if n&gt;\epsilon, then \frac{1}{n} &lt; \frac{1}{\epsilon}, and 2^{\frac{1}{n}} &lt; 2^{\frac{1}{\epsilon}}

 

[VietDao29]

if n&gt;\epsilon, then \frac{1}{n} &lt; \frac{1}{\epsilon}, and 2^{\frac{1}{n}} &lt; 2^{\frac{1}{\epsilon}}

Well, and how can you go from there?

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

Homework Statement

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

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?

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? :)