Calculate $\lim_{n\to\infty}\frac{e^\frac{1}{n}+e^\frac{2}{n}+\ldots+e}{n}$ Calculate $$\lim_{n\to\infty}\frac{e^\frac{1}{n}+e^\frac{2}{n}+\ldots+e^\frac{n-1}{n}+e}{n}$$by expressing this limit as a definite integral of some continuous function and then using calculus methods.

I've worked this out with a friend and we've come to the conclusion that this is equivalent to $\int e^\frac{x}{n}\,dx$. However, we would like a confirmation that this is what the above evaluates to. 
 A: If $f(x)=e^x$ then the expression is $\frac{1}{n}\sum_{k=1}^nf(\frac{k}{n})$, which is a right Riemann sum for $\int_0^1f(x)\;dx$.
Therefore the limit is $\int_0^1e^x\;dx=e-1$.
A: If you do not use the integral (as commented by @carmichael561), you coud write $$S_n=\frac{e^\frac{1}{n}+e^\frac{2}{n}+\ldots+e^\frac{n-1}{n}+e}{n}=\frac{\sum_{i=1}^n e^{\frac in} }n=\frac{\sum_{i=1}^n x^i }n$$ using $x= e^{\frac 1n}$.
Hence $$S_n=\frac{x \left(1-x^n\right)}{n(1-x)}=\frac{(e-1) e^{\frac{1}{n}}}{\left(e^{\frac{1}{n}}-1\right) n}$$ Now, for large values of $n$, you could use Taylor expansion $$e^{\frac{1}{n}}=1+\frac{1}{n}+\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ which makes $$S_n=\frac{e-1}n \times\frac {1+\frac{1}{n}+\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)}{\frac{1}{n}+\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)}=\frac{e-1}n \times\left( n+\frac{1}{2}+O\left(\frac{1}{n}\right)\right)$$ $$=(e-1)+\frac{(e-1)}{2 n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approched,
