Answer
$\sum ^{n}_{k=1}k^{-\frac {3}{2}}$
Work Step by Step
$\dfrac {\sqrt {1}}{1^{2}}+\dfrac {\sqrt {2}}{2^{2}}+\dfrac {\sqrt {3}}{3^{2}}+\ldots +\dfrac {\sqrt {n}}{n^{2}}=\sum ^{n}_{k=1}\dfrac {\sqrt {k}}{k^{2}}=\sum ^{n}_{k=1}k^{-\frac {3}{2}}$
