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?
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.
As an alternative note that by geometric series
$$\frac{1}{n}\sum_{k=1}^{n} e^{{5k}/{n}}=\frac{1}{n}\sum_{k=0}^{n-1} (e^{5/n})^{k+1}=\frac{e^{5/n}}{n}\sum_{k=0}^{n-1} (e^{5/n})^{k}=\frac{e^{5/n}}{n}\frac{e^5-1}{e^{5/n}-1}=e^{5/n}\frac15\frac{5/n}{e^{5/n}-1}(e^5-1)\\\to\frac{e^5-1}{5}$$
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 $46.6808$.
