Answer
$\displaystyle \sum_{k=1}^{n}\frac{k}{e^{k}}$
Work Step by Step
There are $n$ terms ( as the numerators go from 1 to n.)
Count them with index k: $\displaystyle \sum_{k=1}^{n}$(...)
The first term (when k=$1$) is $\displaystyle \frac{1}{e}=\frac{1}{e^{1}}$,
The second term (when k=$2$) is $\displaystyle \frac{2}{e^{2}}$,
The third term (when k=$3$) is $\displaystyle \frac{3}{e^{3}}$,
The kth is $\displaystyle \frac{k}{e^{k}}$ so
$\displaystyle \frac{1}{e}+\frac{2}{e^{2}}+\frac{3}{e^{3}}+\dots+\frac{n}{e^{n}}=\sum_{k=1}^{n}\frac{k}{e^{k}}$
