We’ll need to compute \(L\).

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

When computing \({a_{n + 1}}\) be careful to pay attention to any coefficients of \(n\) and powers of \(n\). Failure to properly deal with these is one of the biggest mistakes that students make in this computation and mistakes at that level often lead to the wrong answer!

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