$\lim_{n\to \infty}\sum_{r = 1}^{n}1 - \frac{1}{n} e^{r}$ is
Ae
B$\mathrm{e} - 1$
C$1 - \mathrm{e}$
D$\mathrm{e} + 1$
Answer & Solution
Correct answer: B. $\mathrm{e} - 1$
Interpreting the given expression in the standard Riemann-sum form,
$$\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n} e^{r/n}.$$
This matches
$$\int_0^1 e^x\,dx.$$
Now,
$$\int_0^1 e^x\,dx = \left[e^x\right]_0^1 = e-1.$$
Comparing with the options, this is option $\mathrm{(B)}$.
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