Question

1. -/1.11 points determine whether the sequence converges or diverges. if it converges, find the limit. (if the sequence diverges, enter diverges.) {\frac{(6n - 1)!}{(6n + 1)!}} \lim_{n\to\infty} a_n = 12. -/1.11 points determine whether the sequence converges or diverges. if it converges, find the limit. (if the sequence diverges, enter diverges.) {\frac{\ln(n)}{\ln(3n)}} \lim_{n\to\infty} a_n =

Explanation:

Step1: Simplify the factorial expression

Use the property \(n! = n\times(n - 1)\times\cdots\times1\). So, \((6n+1)!=(6n + 1)\times(6n)\times(6n-1)!\). Then \(\frac{(6n-1)!}{(6n + 1)!}=\frac{(6n-1)!}{(6n+1)\times(6n)\times(6n-1)!}\). Cancel out \((6n - 1)!\) terms. We get \(\frac{1}{(6n+1)\times(6n)}\).

Step2: Find the limit

Now, find \(\lim_{n
ightarrow\infty}a_{n}=\lim_{n
ightarrow\infty}\frac{1}{(6n+1)\times(6n)}\). Expand the denominator: \((6n+1)\times(6n)=36n^{2}+6n\). Then \(\lim_{n
ightarrow\infty}\frac{1}{36n^{2}+6n}\). As \(n
ightarrow\infty\), for the function \(y = 36n^{2}+6n\), when \(n\) is very large, \(36n^{2}+6n\approx36n^{2}\). And \(\lim_{n
ightarrow\infty}\frac{1}{36n^{2}+6n}=\lim_{n
ightarrow\infty}\frac{1}{n^{2}(36+\frac{6}{n})}\). Since \(\lim_{n
ightarrow\infty}\frac{1}{n}=0\), we know that \(\lim_{n
ightarrow\infty}\frac{1}{n^{2}(36+\frac{6}{n})}=0\).

Answer:

\(0\)