Answer
The series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{\cos \left( {\dfrac{\pi }{4} + 2\pi n} \right)}}{{\sqrt n }}$ diverges.
Work Step by Step
We have $\cos \left( {\dfrac{\pi }{4} + 2\pi n} \right) = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$, ${\ \ \ }$ for $n = 1,2,3,\cdot\cdot\cdot$.
Thus, we can write the original series as
$\mathop \sum \limits_{n = 1}^\infty \dfrac{{\cos \left( {\dfrac{\pi }{4} + 2\pi n} \right)}}{{\sqrt n }} = \dfrac{1}{{\sqrt 2 }}\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{\sqrt n }}$
The series $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{{\sqrt n }}$ diverges by Theorem 3 of Section 11.3, that is, by the Convergence of $p$-Series (because $p = \dfrac{1}{2} \lt 1$).
Therefore, the series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{\cos \left( {\dfrac{\pi }{4} + 2\pi n} \right)}}{{\sqrt n }}$ diverges.
