Biz \(L\) ni hisoblashimiz kerak.

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{a_{n + 1}}\frac{1}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\bf{e}}^{4\left( {n + 1} \right)}}}}{{\left( {n + 1 - 2} \right)!}}\frac{{\left( {n - 2} \right)!}}{{{{\bf{e}}^{4n}}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\bf{e}}^{4n + 4}}}}{{\left( {n - 1} \right)!}}\frac{{\left( {n - 2} \right)!}}{{{{\bf{e}}^{4n}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\bf{e}}^{4n + 4}}}}{{\left( {n - 1} \right)\left( {n - 2} \right)!}}\frac{{\left( {n - 2} \right)!}}{{{{\bf{e}}^{4n}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\bf{e}}^4}}}{{n - 1}}} \right| = 0\end{align*}\]

\({a_{n + 1}}\) 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 \(L = 0 < 1\) va shunga o'xshash seriyani nisbati bilan ko'rishimiz mumkin birlashadi.