Answer
$\sum_{k=1}^{n}\frac{\sqrt{k}}{k^{2}}$
Work Step by Step
We write the sum in sigma notation:
(We notice that the numerator takes the form $\sqrt{k}$, while the denominator takes the form $k^2$, with $k$ being consecutive integers.)
$\frac{\sqrt{1}}{1^{2}}+\frac{\sqrt{2}}{2^{2}}+\frac{\sqrt{3}}{3^{2}}+\cdots+\frac{\sqrt{n}}{n^{2}}=\sum_{k=1}^{n}\frac{\sqrt{k}}{k^{2}}$