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$\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n} e^{\frac{-}{n}}$ is
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The correct answer is:
$e-1$
$e-1$
$\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n} e^{\frac{r}{n}}=\int_0^1 e^x d x=(e-1)$
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