How to integrate $xe^x$ without using antiderivatives or integration by parts.
We have
$$U(f,P_n) = \frac1{n^2} \sum_{k=1}^n k e^{k/n}= L(f,P_n) + \frac{e}{n}.$$
If the limit of the upper sum exists, then it is identical to the limit of the lower sum.
Note that
$$\sum_{k=1}^nk r^k = \frac{r-r^{n+1}}{(1-r)^2}- \frac{nr^{n+1}}{1-r}.$$
Using $r = e^{1/n}$ we have as $n \to \infty$
$$U(f,P_n)= \frac{1/n}{1-e^{1/n}}\frac{1/n}{1-e^{1/n}}e^{1/n}(1-e)- \frac{1/n}{1-e^{1/n}}e^{1/n}e \to 1,$$
since
$$\lim_{n \to \infty} \frac{1/n}{1-e^{1/n}}= -1$$