$$\sum_{k=1}^{n}{e^-k}$$
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MathBot Answer:
\[\sum_{k=1}^{n} e^{- k} = \frac{1 - e^{n}}{\left(1 - e\right) e^{n}}\]
$$\begin{aligned}\sum_{k=1}^{n} e^{- k} &= \sum_{k=0}^{n - 1} e^{- k - 1} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k + p}^{m + p} f{\left(n - p \right)} = \sum_{n=k}^{m} f{\left(n \right)}}\\&= \sum_{k=0}^{n - 1} \frac{1}{e e^{k}}\\&= \frac{\sum_{k=0}^{n - 1} e^{- k}}{e} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=k}^{m} C f{\left(n \right)} = C \sum_{n=k}^{m} f{\left(n \right)}}\\&= \frac{1 - e^{-1}^{1 + n - 1}}{e \left(1 - e^{-1}\right)} \ \ \ \small{\color{grey}\text{by the identity} \sum_{n=0}^{m} a^n = \frac{1-a^{m+1}}{1-a}}\\&= \frac{1 - e^{n}}{\left(1 - e\right) e^{n}}\end{aligned}$$
\(e\) is Euler's number, a mathematical constant that is approximately \(2.71828\).