Find $\lim_{n\rightarrow \infty }\left(\sum_{k=0}^{n-1}{\frac{e^{\frac{k}{n}}}{n}}\right)$. 
Find $$\lim_{n\rightarrow \infty }\left ( \frac{1}{n} + \frac{e^{\frac{1}{n}}}{n} +  \frac{e^{\frac{2}{n}}}{n} + \frac{e^{\frac{3}{n}}}{n}+.....+ \frac{e^{\frac{n-1}{n}}}{n}\right ).$$

Solving a bit and applying GP, I got 
$\left ( e-1 \right )\lim_{n\rightarrow \infty } \frac{1}{n.\left ( e^{\frac{1}{n}} -1 \right )}$
Now, limit gives the expression as 
$\left ( e-1 \right )\lim_{n\rightarrow \infty } \frac{1}{\infty *0}$

How do I find it now? Should I use the $\frac{0}{0}$ form?

 A: You can continue with your way of thinking. Using the geometric progression formula we have that the sum is equal to:
$$\lim_{n \to \infty} \frac{e- 1}{n(e^{\frac 1n}-1)}$$
Now you can switch to real numbers and use the fact that:
$$\lim_{n \to \infty} \frac{1}{n(e^{\frac 1n}-1)} = \lim_{x \to \infty} \frac{1}{\frac{e^{\frac 1x}-1}{\frac 1x}} = \frac{1}{\lim_{x \to \infty} \frac{e^{\frac 1x}-1}{\frac 1x - 0}} = \frac{1}{(e^x)'|_{x=0}} = \frac{1}{e^0} = 1$$
Hence the sum is equal $e-1$
A: Hint: this is the Riemann sum of the following integral:
$$\int_0^1e^x\ dx$$
Which is easily solved to give $e-1$.
A: You were already pretty close with your own way!:
$$\lim_{n\to\infty}\frac1{n\left(e^{1/n}-1\right)}\stackrel{x:=\frac1n}=\lim_{x\to0}\frac x{e^x-1}$$
And now observe that
$$\lim_{x\to0}\frac{e^x-1}x=\left(e^x\right)'|_{x=0}=e^0=1$$
A: @Jon Garrick: just forwarding your answer $$\displaystyle\left ( e-1 \right )\lim_{n\rightarrow \infty } \frac{1}{n.\left ( e^{\frac{1}{n}} -1 \right )}$$
$$\displaystyle\left ( e-1 \right )\lim_{n\rightarrow \infty } \frac{1}{n.\left ( 1+\frac1n+O(\frac1{n^2}) -1 \right )}=\displaystyle\left ( e-1 \right )\lim_{n\rightarrow \infty } \frac{1}{\left ( 1+O(\frac1{n})\right )}=e-1$$
