Ciao,
\[\sum\limits_{n = 1}^{ + \infty } {\frac{n}{{{{\ln }^n}(\frac{{1 + {e^2}n}}{{2 + n}})}}} \]
utilizzando il criterio della radice si ha:
\[\mathop {\lim }\limits_{x \to + \infty } \sqrt[n]{{{a_n}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt[n]{{\frac{n}{{{{\ln }^n}(\frac{{1 + {e^2}n}}{{2 + n}})}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt[n]{n}}}{{\ln (\frac{{1 + {e^2}n}}{{2 + n}})}} \cong \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt[n]{n}}}{{\ln (\frac{{{e^2}n}}{n})}} = \]
\[ = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt[n]{n}}}{{\ln ({e^2})}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt[n]{n}}}{{2\ln (e)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt[n]{n}}}{2} = \frac{1}{2}\mathop {\lim }\limits_{x \to + \infty } \sqrt[n]{n} = \frac{1}{2}\mathop {\lim }\limits_{x \to + \infty } {n^{\frac{1}{n}}} = \]
\[ = \frac{1}{2}\mathop {\lim }\limits_{x \to + \infty } {e^{\ln ({n^{\frac{1}{n}}})}} = \frac{1}{2}\mathop {\lim }\limits_{x \to + \infty } {e^{\frac{1}{n}\ln (n)}} = \frac{1}{2}\mathop {\lim }\limits_{x \to + \infty } {e^{\frac{{\ln (n)}}{n}}} = \frac{1}{2}{e^0} = \frac{1}{2} < 1\]
\[ \to \;\;\;\sum {converge\;!} \]
Saluti.
