Biz \(L\) ni hisoblashimiz kerak.

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{a_{n + 1}}\frac{1}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {2\left( {n + 1} \right)} \right)!}}{{5\left( {n + 1} \right) + 1}}\frac{{5n + 1}}{{\left( {2n} \right)!}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {2n + 2} \right)!}}{{5n + 6}}\frac{{5n + 1}}{{\left( {2n} \right)!}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {2n + 2} \right)\left( {2n + 1} \right)\left( {2n} \right)!}}{{5n + 6}}\frac{{5n + 1}}{{\left( {2n} \right)!}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {2n + 2} \right)\left( {2n + 1} \right)\left( {5n + 1} \right)}}{{5n + 6}}} \right| = \infty \end{align*}\]

\({a_{n + 1}}\) 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 \(L = \infty > 1\) va shunga o'xshash seriyani nisbati bilan ko'rishimiz mumkin farqlanadi.