Biz \(L\) ni hisoblashimiz kerak.

\[L = \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{\left| {{a_n}} \right|}} = \mathop {\lim }\limits_{n \to \infty } {\left| {\frac{{{n^{1 - 3n}}}}{{{4^{2n}}}}} \right|^{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^{\frac{1}{n} - 3}}}}{{{4^2}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^{\frac{1}{n}}}\,\,{n^{ - 3}}}}{{{4^2}}}} \right| = \frac{{\left( 1 \right)\left( 0 \right)}}{{16}} = 0\]

2-qadam

OK, biz \(L = 0 < 1\) va shunga o'xshash seriyani Root Test orqali ko'rishimiz mumkin birlashadi.