TITLE: Find the limit of the given sequence
QUESTION [5 upvotes]: $$\lim_{n\to \infty}\sqrt[n]{1^n+2^n+3^n+\dots +2012^n}$$
I have used the Squeeze theorem (or the Sandwich theorem):
$$n\cdot \sqrt[n]{1^n}\le\sqrt[n]{1^n+2^n+3^n+\dots +2012^n}\le n\cdot \sqrt[n]{2012^n}$$
$\lim\limits_{n\to \infty} n\cdot \sqrt[n]{1^n} = \infty$
$\lim\limits_{n\to \infty} n\cdot \sqrt[n]{2012^n} = \infty$
thus the limit of that sequence equals $\infty$.
Is that correct?

REPLY [3 votes]: To supplement the above answers - we can generalize the obtained result. If $k$ is a natural number, and $0\leq x_{1}\leq x_{2}\leq ...\leq x_{k}$, then
$$\lim_{n\to\infty}\sqrt[n]{x_{1}^{n}+x_{2}^{n}+...+x_{k}^{n}}=x_{k}.$$
Indeed, the following inequalities hold:
$$x_{k}=\sqrt[n]{x_{k}^{n}}\leq\sqrt[n]{x_{1}^{n}+x_{2}^{n}+...+x_{k}^{n}}\leq\sqrt[n]{k\cdot x_{k}^{n}}=x_{k}\cdot\sqrt[n]{k}.$$
Since we know that $\sqrt[n]{k}\to 1$, our result follows from the squeeze theorem.