How can I calculate this Riemann sum?

Question

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?

Answer 1

$$\lim_{n\to\infty} \sum_{k=1}^{n-1}\frac{e^{\frac{k}{n}}}{n}=\int_0^1e^xdx=e-1$$

Answer 2

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$.

Answer 3

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 $1.552181$.