Solving $\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n} e^{{5k}/{n}}$ This was a question on one of my exam practice problems and although I know the solution, I do not see how to get to it. 
Given the question 
$$\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n} e^{{5k}/{n}}$$
I can see the $e^{{5k}/{n}}$ becoming some form of $e^5$ but I don't really know how to show this nor come to the correct answer. Are there any formulas that can help solve these problems? 
 A: This is the (right) Riemann sum of the function $f(x) = e^{5x}$, so 
$$
\lim_{n  \to \infty}\frac{1}{n} \sum_{k = 1}^n e^{\frac{5k}{n}} = \int_0^1 e^{5x}\, dx.
$$
Then evaluate this integral.
A: As an alternative note that by geometric series
$$\frac{1}{n}\sum_{k=1}^{n} e^{{5k}/{n}}=\frac{1}{n}\sum_{k=0}^{n-1} (e^{5/n})^{k+1}=\frac{e^{5/n}}{n}\sum_{k=0}^{n-1} (e^{5/n})^{k}=\frac{e^{5/n}}{n}\frac{e^5-1}{e^{5/n}-1}=e^{5/n}\frac15\frac{5/n}{e^{5/n}-1}(e^5-1)\\\to\frac{e^5-1}{5}$$
A: In the same spirit as gimusi in his/her answer
$$S_n=\frac{1}{n}\sum_{k=1}^{n} e^{{ak}/{n}}=\frac{\left(e^a-1\right) e^{a/n}}{n \left(e^{a/n}-1\right)}$$ Now, using Taylor expansion for large values of $n$ will lead to
$$S_n=(e^a-1)\left(\frac{1}{a}+\frac{1}{2 n}+\frac{a}{12 n^2}+O\left(\frac{1}{n^3}\right) \right)$$ which shows the limit and how it is approched.
Moerover, it allows pretty accurate estimates of the partial sums.
For you case where $a=5$, $S_5=\frac{1}{5} e \left(1+e+e^2+e^3+e^4\right)\approx 46.6408$ while the above expansion would give $46.6808$.
