$\lim_{n \to \infty} \frac 1n(\sqrt[n]{e}+\sqrt[n]{e^2}+...+\sqrt[n]{e^{2n}})$ $\lim_{n \to \infty} \frac 1n(\sqrt[n]{e}+\sqrt[n]{e^2}+...+\sqrt[n]{e^{2n}})$
This problem is from Spivak 1994. I can see that it is integral sum for $e^x$ for $x$ from $0$ to $2$. So when I calculated this I got $\frac12\int e^xdx$, however in the answers it is given $\int e^xdx$ without $\frac 12$. I think this is not right because we are itegrating from 0 to 2 and the length of each interval is $ \frac 2n$. So we need to write $\lim_{n \to \infty} \frac 1n(\sqrt[n]{e}+\sqrt[n]{e^2}+...+\sqrt[n]{e^{2n}})$ as $\frac 12 \lim_{n \to \infty} \frac 2n(\sqrt[n]{e}+\sqrt[n]{e^2}+...+\sqrt[n]{e^{2n}})$
And only then we can write $\lim_{n \to \infty} \frac 2n(\sqrt[n]{e}+\sqrt[n]{e^2}+...+\sqrt[n]{e^{2n}})$ as  $\int e^xdx$
My question: I dont need the solution. I can see that it is integral sum. What I dont know is should there be 1/2 or not. I explained in the body why I think there should be 1/2 before integral
I understood that there are 2n terms, and if we integrate from 0 to 2, length of each interval will be 1/n. How can I close the question?
 A: Your sum can be rewritten as
$$
\frac{1}{n}\sum\limits_{k=1}^{2n} e^{k/n} = \frac{1}{n}\sum\limits_{k=1}^{2n} f\left(\frac{k}{n}\right)
$$
where $f(x) = e^x$. Does this answer your question?
Edit: Note that I changed the sum to go up to $2n$, but the equations you wrote were inconsistent since you also wrote "So we need to write $\lim_{n\to \infty} \frac{1}{n}\left(\sqrt[n]{e} + \cdots + \sqrt[n]{e^n}\right)$..." which implies the sum goes to $n$ (not $2n$). 
A: I understood that there are 2n terms, and if we integrate from 0 to 2, length of each interval will be 1/n. 
A: With $r=e^{1/n}$ the geometric series :
$$\frac1n\sum_{k=1}^{2n}e^{k/n}=\frac1n\sum_{k=1}^{2n}r^k=\frac1n\frac{1-r^{2n}}{1-r}=\frac1n\frac{1-e^{2}}{1-e^{1/n}}=\frac{e^{2}-1}{n(e^{1/n}-1)}$$
$n\to\infty\quad\implies\quad \frac{1}{n}\to 0\quad;\quad e^{1/n}\sim 1+\frac1n+O\left(\frac{1}{n^2}\right)$
$ n(e^{1/n}-1)\sim n\left(1+\frac1n+O\left(\frac{1}{n^2}\right)-1\right)\sim 1+O\left(\frac{1}{n}\right)$
$\implies\quad\lim_{n\to\infty} n(e^{1/n}-1)=1$
$$\lim_{n\to\infty}\frac1n\sum_{k=1}^{2n}e^{k/n}=\frac{e^{2}-1}{1}=e^2-1$$
$$ $$Alternatively, integration method :
$$\frac1n\int_{2}^{2n}e^{k/n}dk<\frac1n\sum_{k=1}^{2n}e^{k/n}<\frac1n\int_{1}^{2n-1}e^{k/n}dk$$
$$e^2-e^{2/n}<\frac1n\sum_{k=1}^{2n}e^{k/n}<e^{2-\frac1n}-e^{1/n}$$
For $n\to\infty$ , both $\quad e^2-e^{2/n}\quad$ and $\quad e^{2-\frac1n}-e^{1/n}\quad$ tend to $\quad e^2-1,\quad$thus :
$$\lim_{n\to\infty}\frac1n\sum_{k=1}^{2n}e^{k/n}=e^2-1$$
