Which sum converts into a definite integral? $\sum_{i=0}^n \frac {e ^ i}{n}$ or $\sum_{i=0}^n \frac {e ^ \frac{i}{n}}{n}$

Question

Which sum converts into a definite integral? $\sum_{i=0}^n \frac {e ^ i}{n}$ or $\sum_{i=0}^n \frac {e ^ \frac{i}{n}}{n}$ Or none of the two do?

I have come across problems such as $ \lim_{n_\to infinity} (\frac{n!}{n^n})^\frac{1}{n}$ . When we take a log of this limit, the product of n! and n.n.n. ... n turns into a sum, while the exponent 1/n becomes a common term for the entire sum. This expression now becomes a definite integral from 0 to 1, as explained in https://www.youtube.com/watch?v=89d5f8WUf1Y.

I suspect that one of the two above mentioned sums, because of the presence of 1/n as a term, can turn into a definite integral. Is this true? If so, is it the sum with $e^i$ or $e^\frac{i}{n}$?

I don't think $e^i$ sum can converge, since the sum is essentially a sum of a GP, and the $\frac{(e)\frac{(e^{n} - 1 )}{e - 1}}{n}$ tends to infinity as n goes to infinity. On the other hand, with $e^{i/n}$ sum, it turns into $\frac{(e)\frac{(e - 1 )}{e ^\frac{1}{n} - 1}}{n}$, which I think simplifies to e. (e - 1)

As I type this, I am also getting a feeling that I have maybe mixed up definite integrals with a geometric series. Can someone elaborate here and clarify?

My questions - restated - is, which of the two given sums can be converted into a definite integral? First, second or neither?

Answer 1

Let $P_n= \{0, \frac{1}{n}, \frac{2}{n},..., \frac{n}{n} \}$ and $f(x)=e^x.$

Then

$$ L(P_n,f)=\sum_{i=0}^n \frac {e ^ \frac{i}{n}}{n}.$$

Can you proceed ?