Answer
False.
Work Step by Step
False.
As a counterexample, consider ${u_k} = \dfrac{1}{{{k^{3/4}}}}$.
Thus,
$\sum {\left( {{u_k}} \right)^2} = \sum \dfrac{1}{{{k^{3/2}}}}$
The series $\sum {\left( {{u_k}} \right)^2} = \sum \dfrac{1}{{{k^{3/2}}}}$ converges because it is a convergent $p$-series with $p = \dfrac{3}{2} \gt 1$. However, the $p$-series $\sum {u_k} = \sum \dfrac{1}{{{k^{3/4}}}}$ diverges ($p = \dfrac{3}{4} \lt 1$).
