Convergence of $\frac{1}{2n}\sqrt[n]{1^n+2^n+...+(2n)^n}$? $a_n=\frac{1}{2n}\sqrt[n]{1^n+2^n+...+(2n)^n}$
I think that $\sqrt[n]{1^n+2^n+...+(2n)^n}\rightarrow 2n+1$
So $a_n=\frac{2n+1}{2n}\rightarrow 1$
Any ideas if that's correct? And if so how do I prove it?
 A: Use the fact that
$$
(2n)^n \leq
1^n+2^n+...+(2n)^n \leq
2n(2n)^n
$$
and the squeeze theorem.
A: Hint.  For $j,n\in \Bbb Z^+$ we have      $$\quad\int_{j-1}^jx^ndx<j^n<\int_j^{j+1}x^ndx.$$ Now sum from $j=1$ to $j=2n.$
A: We have 
\begin{align}
a_n=&
\frac{1}{2n}\sqrt[n]{1^n+2^n+...+(2n)^n}
\\
=&\frac{1}{2n}\sqrt[n]{(2n)^n\left[\frac{1^n}{(2n)^{n}}+\frac{2^n}{(2n)^{n}}+...+\frac{(2n-1)^n}{(2n)^{n}}+1\right]}
\\
=&\sqrt[n]{\frac{1^n}{(2n)^{n}}+\frac{2^n}{(2n)^{n}}+...+\frac{(2n-1)^n}{(2n)^{n}}+1}
\\
=&\sqrt[n]{\left(\frac{1}{2n}\right)^{n}+\left(\frac{2}{2n}\right)^{n}+...+\left(1-\frac{1}{2n}\right)^{n}+1}
\end{align}
For $ n $ big enough we have
$$
0\leq \left(\frac{1}{2n}\right)^{n}\leq 1, \quad 0\leq \left(\frac{2}{2n}\right)^{n} \leq 1,\ldots \qquad \ldots, 0\leq \left(1-\frac{1}{2n}\right)^{n}\leq 1.
$$
It is easy to see that
$$
\sqrt[n]{0+0+\ldots+0+1}\leq a_n \leq \sqrt[n\,]{\underbrace{1+1+\ldots+1+1}_{n \mbox{ times }}}
$$
or 
$$
1\leq a_n\leq \sqrt[n]{n}
$$
Once $\sqrt[n]{n}\to 1$, thus $a_n\to 1$.
