Answer
The series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{\cos \left( {\dfrac{\pi }{4} + \pi n} \right)}}{{\sqrt n }}$ converges conditionally.
Work Step by Step
1. Since $ - 1 \le \cos \left( {\dfrac{\pi }{4} + \pi n} \right) \le 1$, so the following inequality is valid:
$\left| {\dfrac{{\cos \left( {\dfrac{\pi }{4} + \pi n} \right)}}{{\sqrt n }}} \right| \ge \dfrac{1}{n}$
We know that the harmonic series $\mathop \sum \limits_{n = 1}^\infty \dfrac{1}{n}$ diverges. By the Direct Comparison Test, the larger series $\mathop \sum \limits_{n = 1}^\infty \left| {\dfrac{{\cos \left( {\dfrac{\pi }{4} + \pi n} \right)}}{{\sqrt n }}} \right|$ also diverges.
2. Consider the term $\cos \left( {\dfrac{\pi }{4} + \pi n} \right)$
We have
$\cos \left( {\dfrac{\pi }{4} + \pi n} \right) = \left\{ {\begin{array}{*{20}{c}}
{ - \dfrac{1}{{\sqrt 2 }},}&{n = 1,3,5,\cdot\cdot\cdot}\\
{\dfrac{1}{{\sqrt 2 }},}&{n = 2,4,6,\cdot\cdot\cdot}
\end{array}} \right.$
Thus, we can write the original series as
$\mathop \sum \limits_{n = 1}^\infty \dfrac{{\cos \left( {\dfrac{\pi }{4} + \pi n} \right)}}{{\sqrt n }} = \mathop \sum \limits_{n = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^n}}}{{\sqrt {2n} }}$
The terms ${b_n} = \dfrac{1}{{\sqrt {2n} }}$ are positive, decreasing and converges to $0$. Therefore, $\mathop \sum \limits_{n = 1}^\infty \dfrac{{\cos \left( {\dfrac{\pi }{4} + \pi n} \right)}}{{\sqrt n }}$ converges by Theorem 2 of Section 11.4, that is, by the Alternating Series Test.
Hence, by definition, the series $\mathop \sum \limits_{n = 1}^\infty \dfrac{{\cos \left( {\dfrac{\pi }{4} + \pi n} \right)}}{{\sqrt n }}$ converges conditionally.
