Limit series $(a^n-b^n)^{1/n}$ I have problem with calculate limit of $(a^n-b^n)^{1/n}$ where $b>a$
When $a>b$ it's easier. Is true that $\lim_{n \to\infty} (-1)^{1/n}=1$?
If so how prove it. I think that need to take the proof in complex number.
 A: Over the reals, we don't define $x^a$ unless $x$ is nonnegative or $a$ is an integer. Over the complexes, there are several choices, and one is forced to make a specific choice to make the limit make sense.
If $a=x^n$ (so $x$ is a candidate for $x=a^{1/n}$), then so is $xe^{2\pi ik/n}$ for any $k\in\Bbb Z$:
$$(xe^{2\pi ik/n})^n=x^n\cdot e^{2\pi ik}=a\cdot1=a.$$
Since $xe^{2\pi ik/n}=xe^{2\pi im/n}$ whenever $k-m$ is a multiple of $n$, there are really $n$ different choices here. The "canonical" choice is $x=e^{\frac1n{\log a}}$, where we choose the branch of $\log a$ so that $\Im[\log a]\in(-\pi,\pi]$. For this choice:
$$\lim_{n\to\infty}(-1)^{1/n}=\lim_{n\to\infty}e^{1/n\log(-1)}=\lim_{n\to\infty}e^{\pi i/n}=\lim_{n\to\infty}\cos\frac{\pi}n+i\sin\frac{\pi}n=1$$
For the general case (when $0<a<b$), note that $(b^n-a^n)^{1/n}=b(1-(a/b)^n)^{1/n}$ is bounded by $b$ above and $b(1-(a/b)^n)$ below (since $x^n\le x\Rightarrow x\le x^{1/n}$ for all $0<x<1$). Now $(a^n-b^n)^{1/n}=(b^n-a^n)^{1/n}(-1)^{1/n}$, so the real part of $\lim_{n\to\infty}(a^n-b^n)^{1/n}$ is 
$$\begin{align}
\lim_{n\to\infty}(a^n-b^n)^{1/n}&=\lim_{n\to\infty}(b^n-a^n)^{1/n}(-1)^{1/n}\\
&=\lim_{n\to\infty}(b^n-a^n)^{1/n}\lim_{n\to\infty}(-1)^{1/n}\\
&=\lim_{n\to\infty}(b^n-a^n)^{1/n}\\
&\ge b\lim_{n\to\infty}(1-(a/b)^n)=b\end{align}$$
and it is also at most $b$ because of the upper bound. Thus $\lim_{n\to\infty}(a^n-b^n)^{1/n}=b$ for $0<a<b$, and by switching $a$ and $b$ in the proof above, $\lim_{n\to\infty}(a^n-b^n)^{1/n}=a$ when $0<b<a$.
A: We don't define roots of negative numbers, hence I suppose your problem already assumes $a \geq b$
Or well, OK, we define them since you've already mentioned complex numbers, but then they aren't 'real' functions, they're multivalued so the question if there's a limit would be nonsense.
