Ciao,
\[\sum\limits_{n = 1}^{ + \infty } {n\,{{(\frac{1}{n}\left| {\cos (n)} \right|)}^n}} \]
applicando il criterio della radice si ha
\[\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\,{{(\frac{1}{n}\left| {\cos (n)} \right|)}^n}}} = \mathop {\lim }\limits_{n \to + \infty } (\frac{1}{n}\left| {\cos (n)} \right|)\sqrt[n]{{n\,}} = \mathop {\lim }\limits_{n \to + \infty } (\frac{1}{n}\left| {\cos (n)} \right|) \cdot \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\,}} = \]
\[ = \mathop {\lim }\limits_{n \to + \infty } (\frac{1}{n}\left| {\cos (n)} \right|) \cdot \mathop {\lim }\limits_{n \to + \infty } {n^{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to + \infty } (\frac{1}{n}\left| {\cos (n)} \right|) \cdot \mathop {\lim }\limits_{n \to + \infty } {e^{\ln ({n^{\frac{1}{n}}})}} = \]
\[ = \mathop {\lim }\limits_{n \to + \infty } (\frac{1}{n}\left| {\cos (n)} \right|) \cdot \mathop {\lim }\limits_{n \to + \infty } {e^{\frac{1}{n}\ln (n)}} = \mathop {\lim }\limits_{n \to + \infty } (\frac{1}{n}\left| {\cos (n)} \right|) \cdot \mathop {\lim }\limits_{n \to + \infty } {e^{\frac{{\ln (n)}}{n}}} = \]
\[ = \mathop {\lim }\limits_{n \to + \infty } (\frac{1}{n}\left| {\cos (n)} \right|) \cdot \mathop {\lim }\limits_{n \to + \infty } {e^{\frac{{\ln (n)}}{n}}} = 0 \cdot {e^0} = 0 \cdot 1 = 0 < 1\;\; \to \;\;converge!\]
Saluti.
