Biz \(L\) ni hisoblashimiz kerak.

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{a_{n + 1}}\frac{1}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{3^{1 - 2\left( {n + 1} \right)}}}}{{{{\left( {n + 1} \right)}^2} + 1}}\frac{{{n^2} + 1}}{{{3^{1 - 2n}}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{3^{ - 1 - 2n}}}}{{{{\left( {n + 1} \right)}^2} + 1}}\frac{{{n^2} + 1}}{{{3^{1 - 2n}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{1}{{{{\left( {n + 1} \right)}^2} + 1}}\frac{{{n^2} + 1}}{{{3^2}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^2} + 1}}{{9\left[ {{{\left( {n + 1} \right)}^2} + 1} \right]}}} \right| = \frac{1}{9}\end{align*}\]

\({a_{n + 1}}\) 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 \(L = \frac{1}{9} < 1\) va shunga o'xshash seriyani nisbati bilan ko'rishimiz mumkin birlashadi.