Question

1. -/1.11 points determine whether the sequence converges or diverges. if it converges, find the limit. (if the sequence diverges, enter diverges.) \\( \left\\{ \frac{\ln (n)}{\ln (3 n)}\

ight\\} \\) \\( \lim _{n \
ightarrow \infty} a_{n}= \\) 13. -/1.11 points determine whether the sequence converges or diverges. if it converges, find the limit. (if the sequence diverges, enter diverges.) \\( \left\\{n^{2} e^{-3 n}\
ight\\} \\) \\( \lim _{n \
ightarrow \infty} a_{n}= \\)

Explanation:

Step1: Use logarithm properties

Use the property \(\ln(ab)=\ln a+\ln b\). So, \(\ln(3n)=\ln3+\ln n\). Then \(a_{n}=\frac{\ln n}{\ln(3n)}=\frac{\ln n}{\ln3 + \ln n}\).

Step2: Divide numerator and denominator by \(\ln n\)

Let \(t = \ln n\) (as \(n\to\infty\), \(t\to\infty\)). Then \(a_{n}=\frac{t}{\ln3 + t}=\frac{1}{\frac{\ln3}{t}+1}\).

Step3: Find the limit as \(n\to\infty\)

As \(n\to\infty\), \(t=\ln n\to\infty\). Then \(\lim_{n\to\infty}a_{n}=\lim_{t\to\infty}\frac{1}{\frac{\ln3}{t}+1}\). Since \(\lim_{t\to\infty}\frac{\ln3}{t}=0\), we have \(\lim_{n\to\infty}a_{n}=1\).

Answer:

\(1\)