Biz \(L\) ni hisoblashimiz kerak.

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{a_{n + 1}}\frac{1}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 2} \right)}^{1 + 3\left( {n + 1} \right)}}\left( {n + 1 + 1} \right)}}{{{{\left( {n + 1} \right)}^2}{5^{1 + n + 1}}}}\frac{{{n^2}{5^{1 + n}}}}{{{{\left( { - 2} \right)}^{1 + 3n}}\left( {n + 1} \right)}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 2} \right)}^{4 + 3n}}\left( {n + 2} \right)}}{{{{\left( {n + 1} \right)}^2}{5^{2 + n}}}}\frac{{{n^2}{5^{1 + n}}}}{{{{\left( { - 2} \right)}^{1 + 3n}}\left( {n + 1} \right)}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 2} \right)}^3}\left( {n + 2} \right)}}{{{{\left( {n + 1} \right)}^2}\left( 5 \right)}}\frac{{{n^2}}}{{\left( {n + 1} \right)}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{ - 8{n^2}\left( {n + 2} \right)}}{{5{{\left( {n + 1} \right)}^2}\left( {n + 1} \right)}}} \right| = \frac{8}{5}\end{align*}\]

\({a_{n + 1}}\) ni hisoblashda \(n\) koeffitsientlariga va \(n\) quvvatlariga e'tibor bering. Bular bilan to'g'ri shug'ullanmaslik talabalarning ushbu hisoblashda eng katta xatolaridan biridir va bu darajadagi xatolar ko'pincha noto'g'ri javobga olib keladi!

2-qadam

OK, biz \(L = \frac{8}{5} > 1\) va shunga o'xshash seriyani nisbati bilan ko'rishimiz mumkin farqlanadi.