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?

• I guess my first question is how did you get $2n+1$? Sep 16 '18 at 16:59
• Correct. By the way, your answer is the only one that show the FINAL RESULT $\displaystyle\left( one, 1\right)$. Sep 16 '18 at 17:06

Use the fact that $$(2n)^n \leq 1^n+2^n+...+(2n)^n \leq 2n(2n)^n$$ and the squeeze theorem.
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 $j=2n.$
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 $a_n\to 1$.