${lim}_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} e^{\frac{r}{n}}$ is

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e - 1

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1 - e

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1 + e

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Solution

The correct option is B e - 1
${lim}_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} e^{\frac{r}{n}} = \int_{0}^{1} e^{x} d x = [e^{x}]_{0}^{1} = e - 1$

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