How can I calculate this Riemann sum? How can I calculate this: $\lim\limits_{n \rightarrow \infty} \frac{1+\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}  } {n}$?
I how come so far:
$$
\lim_{n \rightarrow \infty} \frac{1+\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}  } {n} = \lim_{n \rightarrow \infty} \frac{1}{n} \left(1 +\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}\right) =
$$
$$ \lim_{n \rightarrow \infty}\frac{1}{n}\sum_{i =1}^{n}e^{\frac{i-1}{n}}$$
But how do I continue? 
 A: $$\lim_{n\to\infty} \sum_{k=1}^{n-1}\frac{e^{\frac{k}{n}}}{n}=\int_0^1e^xdx=e-1$$
A: We can avoid integration and use more elementary approach. Note that
$$
\frac{1+\sqrt[n]{e}+\ldots+\sqrt[n]{e^{n-1}}}{n} = \frac{1}{n}\cdot\left(1 +\sqrt[n] {e}+\sqrt[n] {e^2}+ \dots + \sqrt[n] {e^{n-1}}\right) = 
\frac{1}{n}\cdot\frac{(\sqrt[n]{e})^n-1}{\sqrt[n]{e}-1}.
$$
The expression above is equal to $\frac{e-1}{n(\sqrt[n]{e}-1)}$.
Since $\frac{1}{n}\rightarrow 0$ when $n\rightarrow\infty$ and $e^t-1\sim t$ when $t\rightarrow 0$ we obtain
$$
\lim\limits_{n\rightarrow\infty}\frac{e-1}{n(\sqrt[n]{e}-1)}=\lim\limits_{n\rightarrow\infty}\frac{e-1}{n\cdot\frac{1}{n}}=e-1.
$$
Thus, $
\lim\limits_{n\rightarrow\infty}\frac{1+\sqrt[n]{e}+\ldots+\sqrt[n]{e^{n-1}}}{n}=e-1$.
A: If you are concerned by more than the limit itself, consider,as @richrow did,
$$a_n=\frac{e-1}{n(\sqrt[n]{e}-1)}=\frac{e-1}{n(e^{\frac{1}{n}}-1)}$$ and use Taylor expansion to get
$$a_n=\frac{e-1}n \frac 1 {\left(1+\frac{1}{n}+\frac{1}{2 n^2}+\frac{1}{6
   n^3}+O\left(\frac{1}{n^4}\right) \right)-1 }=\frac{e-1}n\left(n-\frac{1}{2}+\frac{1}{12 n}+O\left(\frac{1}{n^2}\right)\right)$$ that is to say
$$a_n=(e-1)\left(1-\frac{1}{2n}+\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)\right)$$
Use it for $n=5$ with your pocket calculator : the exact value would be $1.552177$ while the above truncated series would give $1.552181$.
